Unit 6 – Chapter 10.

Unit 6 – Chapter 10

Unit 6 Chapter 9 Review and Chap. 9 Skills
Section 10.1 – Graph y = ax2 + c Section 10.2 – Graph y = ax2 + bx +c Section 10.3 – Solve Quadratic Eqns by Graphing Section 10.4 – Solve Quad. Eqns. w/Square roots Section 10.6 – Solve Quad. Eqns. With Quad. Formula Section – Compare Linear, Exponential, and Quadratic Models

Warm-Up – Chapter 10

Prerequisite Skills VOCABULARY CHECK Copy and complete the statement. 1. The x-coordinate of a point where a graph crosses the x-axis is a (n) ? ANSWER x-intercept 2. A (n) is a function of the form y = a bx where a 0,b > 0, and b 1. = ? exponential function ANSWER

Prerequisite Skills SKILLS CHECK Draw the blue figure. Then draw its image after a reflection in the red line. 3. ANSWER

Prerequisite Skills SKILLS CHECK Draw the blue figure. Then draw its image after a reflection in the red line. 4. ANSWER

Prerequisite Skills SKILLS CHECK Graph y1 = x and y2 = x – 5. Describe the similarities and differences in the graphs. ANSWER Graphs are Parallel Have the same Slope Y2 is translated Down 5 units

Lesson 10.1, For use with pages 628-634
1. Graph the function y = 2x. ANSWER Identify the domain and range of your graph in Exercise 1. ANSWER domain: all real numbers; range: all positive real numbers

1. Graph the function y = x2. ANSWER Identify the domain and range of your graph in Exercise 1. ANSWER domain: all real numbers; range: all positive real numbers

ANSWER 1. Graph the function y = 3x2. Identify the domain and range of your graph in Exercise 1. ANSWER domain: all real numbers; range: all positive real numbers

1. Graph the function y = (0.1)x2. ANSWER Identify the domain and range of your graph in Exercise 1. ANSWER domain: all real numbers; range: all positive real numbers

ANSWER 1. Graph the function y = x2 + 2 Identify the domain and range of your graph in Exercise 1. ANSWER domain: all real numbers; range: all positive real numbers > 2

1. Graph the function y = (-1)x2 ANSWER Identify the domain and range of your graph in Exercise 1. ANSWER domain: all real numbers; range: all negative numbers

Vocabulary – 10.1 Vertex of a Parabola Quadratic Function
Lowest (or highest) point on a parabola Axis of symmetry Line that passes through the vertex Divides the parabola into two symmetric parts Quadratic Function Non-linear function in the form y = ax2 + bx + c where a ≠ 0 Parabola U – shaped graph of a quadratic function Parent quadratic function Y = x2

Notes – 10.1 – Graph y = ax2 + c All Quadratic Function graphs look like what? A parabola What happens to the shape of the parabola if a >1? It is a vertical stretch It gets skinny! What happens to the shape of a parabola if a < 1? Vertical shrink It gets fat (or “fluffy”!)! What happens if a < 0? The graph curves down. What effect does “c” have on the graph of the function? It translates it up or down.

Examples 10.1

EXAMPLE 1 Graph y= ax2 where a > 1 STEP 1 Make a table of values for y = 3x2 x – 2 – 1 1 2 y 12 3 STEP 2 Plot the points from the table.

EXAMPLE 1 Graph y= ax2 where a > 1 STEP 3 Draw a smooth curve through the points. STEP 4 Compare the graphs of y = 3x2 and y = x2. Both graphs open up and have the same vertex, (0, 0), and axis of symmetry, x = 0. The graph of y = 3x2 is narrower than the graph of y = x2 because the graph of y = 3x2 is a vertical stretch (by a factor of 3) of the graph of y = x2.

x y EXAMPLE 2 Graph y = ax2 where a < 1 Graph y = 1 4 – x2.
Compare the graph with the graph of y = x2. STEP 1 Make a table of values for y = 1 4 – x2. x – 4 – 2 2 4 y – 1

EXAMPLE 2 Graph y = ax2 where a < 1 STEP 2 Plot the points from the table. STEP 3 Draw a smooth curve through the points.

EXAMPLE 2 Graph y = ax2 where a < 1 STEP 4 Compare the graphs of y = 1 4 – x2. and y = x2. Both graphs have the same vertex (0, 0), and the same axis of symmetry, x = 0. However, the graph of 1 4 – x2. y = is wider than the graph of y = x2 and it opens down. This is because the graph of 1 4 – x2. y = is a vertical shrink by a factor of 1 4 with a refl ection in the x-axis of the graph of y = x2.

EXAMPLE 3 Graph y = x2 + c Graph y = x Compare the graph with the graph of y = x2. STEP 1 Make a table of values for y = x2 + 5. x – 2 – 1 1 2 y 9 6 5

EXAMPLE 3 Graph y = x2 + c STEP 2 Plot the points from the table. STEP 3 Draw a smooth curve through the points.

EXAMPLE 3 Graph y = x2 + c STEP 4 Compare the graphs of y = x2 + 5 and y = x2. Both graphs open up and have the same axis of symmetry, x = 0. However, the vertex of the graph of y = x2 + 5, (0, 5), is different than the vertex of the graph of y = x2, (0, 0), because the graph of y = x2 + 5 is a vertical translation (of 5 units up) of the graph of y = x2.

x Y GUIDED PRACTICE for Examples 1, 2 and 3
Graph the function. Compare the graph with the graph of x2. 1. y= –4x2 STEP 1 Make a table of values for y = –4x2 x – 2 – 1 1 2 Y – 16 – 4

GUIDED PRACTICE for Examples 1, 2 and 3 STEP 2 Plot the points from the table. STEP 3 Draw a smooth curve through the points.

x y GUIDED PRACTICE for Examples 1, 2 and 3 2. y = x2 1 3 STEP 1
Make a table of values for y = x2 1 3 x – 6 – 3 3 6 y 12

x Y GUIDED PRACTICE for Examples 1, 2 and 3 3. y = x2 +2 STEP 1
Make a table of values for y = x2 +5 x – 2 – 1 1 2 Y 6 3

EXAMPLE 4 Graph y = ax2 + c Graph y = x2 – 4. Compare the graph with the graph of y = x2. 1 2 STEP 1 Make a table of values for y = x2 – 4. 1 2 x – 4 – 2 2 4 y –2

EXAMPLE 4 Graph y = ax2 + c STEP 2 Plot the points from the table. STEP 3 Draw a smooth curve through the points.

EXAMPLE 4 Graph y = ax2 + c STEP 4 Compare the graphs of y = x2 – 4 and y = x2. Both graphs open up and have the same axis of symmetry, x = 0. However, the graph of y = x2 – 4 is wider and has a lower vertex than the graph of y = x2 because the graph of y = x2 is a vertical shrink and a vertical translation of the graph of y = x2. 1 2

GUIDED PRACTICE for Example 4 Graph the function. Compare the graph with the graph of x2. 4. y= 3x2 – 6 STEP 1 Make a table of values for y = 3x2 – 6. x – 2 – 1 2 1 y 6 – 3 – 6

GUIDED PRACTICE for Example 4 STEP 2 Plot the points from the table. STEP 3 Draw a smooth curve through the points.

GUIDED PRACTICE for Example 4 5. y= –5x2 + 1 STEP 1 Make a table of values for y = –5x2 + 1. x – 2 – 1 1 2 y –19 – 4 –4 – 19

GUIDED PRACTICE for Example 4 6. y = x2 – 2. 3 4 STEP 1 Make a table of values for y = x2 – 2. 3 4 x – 4 – 2 4 2 y 10 1

EXAMPLE 5 Standardized Test Practice How would the graph of the function y = x2 + 6 be affected if the function were changed to y = x2 + 2? A The graph would shift 2 units up. B The graph would shift 4 units up. C The graph would shift 4 units down. D The graph would shift 4 units to the left.

EXAMPLE 5 Standardized Test Practice SOLUTION The vertex of the graph of y = x2 + 6 is 6 units above the origin, or (0, 6). The vertex of the graph of y = x2 + 2 is 2 units above the origin, or (0, 2). Moving the vertex from (0, 6) to (0, 2) translates the graph 4 units down. ANSWER The correct answer is C. A B C D

EXAMPLE 6 Use a graph SOLAR ENERGY A solar trough has a reflective parabolic surface that is used to collect solar energy. The sun’s rays are reflected from the surface toward a pipe that carries water. The heated water produces steam that is used to produce electricity. The graph of the function y = 0.09x2 models the cross section of the reflective surface where x and y are measured in meters. Use the graph to find the domain and range of the function in this situation.

EXAMPLE 6 Use a graph SOLUTION STEP 1
Find the domain. In the graph, the reflective surface extends 5 meters on either side of the origin. So, the domain is 5 ≤ x ≤ 5. STEP 2 Find the range using the fact that the lowest point on the reflective surface is (0, 0) and the highest point, 5, occurs at each end. y = 0.09(5)2 = 2.25 Substitute 5 for x. Then simplify. The range is 0 ≤ y ≤ 2.25.

for Examples 5 and 6 GUIDED PRACTICE Describe how the graph of the function y = x2+2 would be affected if the function were changed to y = x2 – 2. 7. SOLUTION The vertex of the graph of y = x2 + 2 is 6 units above the origin, or (0, 2). The vertex of the graph of y = x2 – 2 is 2 units down above the origin, or (0, – 2). Moving the vertex from (0, 2) to (0, – 2) translates the graph 4 units down. ANSWER The graph would be translated 4 unit down.

for Examples 5 and 6 GUIDED PRACTICE What if In example 6 ,suppose the reflective surface extends just 4 meters on either side of the origin.Find the range of the function in this situation. 8. SOLUTION STEP 1 Find the domain. In the graph, the reflective surface extends 4 meters on either side of the origin. So, the domain is – 4 ≤ x ≤ 4.

for Examples 5 and 6 GUIDED PRACTICE STEP 2
Find the range using the fact that the lowest point on the reflective surface is (0, 0) and the highest point 4, occurs at each end. y = 0.09(4)2 = 144 Substitute 5 for x. Then simplify. The range is – 4 ≤ x ≤ 4, 0 ≤y ≤1.44.

Warm-Up – 10.2

Evaluate the expression. 1. x2 – 2 when x = 3 ANSWER 7 2. 2x2 + 9 when x = 2 ANSWER 17

Evaluate the expression. Martin is replacing a square patch of counter top. The area of the patch is represented by A = s2. What is the area of the patch if the side length is 2.5 inches? 3. ANSWER 6.25 in.2

Vocabulary –10.2 Maximum Value Highest possible value of a graph
Not all graphs have a maximum value Minimum Value Smallest possible value of a graph Not all graphs have a minimum value Vertex AKA the “maximum” or “minimum” value of a parabola

Graph the following equation. 1) Graph y = 3x2 - 12x + 5. 2) Press 2nd  Calc  Minimum 3) Pick two points, one to the left and one to the right of the minimum. 4) What is the X coordinate of that point?? ANSWER X-coordinate is 2 Using the equation y = 3x2 - 12x + 5, calculate the following on your white board. x = - b / (2a) What do you get? Notice anything???? ANSWER X = 2

I bet it only works once!!! 
Lesson 10.2, For use with pages Graph the following equation. What seems to be true about the x coordinate of the vertex (minimum in this case) and the equation x = - b / (2a)? ANSWER They appear to be the same! I bet it only works once!!! 

I bet it only works once!!! 
Lesson 10.2, For use with pages Graph the following equation. What is the line of symmetry for this parabola? Use the Table function to find the y-intercept (Where x = 0). ANSWER 1) x = 2 2) y-intercept (0,5) I bet it only works once!!! 

Notes – 10.2 – Graph ax2 + bx + c PROPERTIES OF GRAPH OF y = ax2 + bx + c If a > 0 opens up and opens down if a < 0 If |a| > 1, gets “skinnier” & if |a| < 1, “fluffier” Vertex has an x-coordinate of – b / (2a) Line of symmetry is x = – b / (2a) Y-intercept of (0,c) TO FIND MINIMUM/MAXIMUM POINTS Find the x-coordinate of the vertex and plug it in to the equation y = ax2 + bx + c

Examples 10.2

EXAMPLE 1 Find the axis of symmetry and the vertex Consider the function y = – 2x2 + 12x – 7. a. Find the axis of symmetry of the graph of the function. b. Find the vertex of the graph of the function.

Find the axis of symmetry and the vertex
EXAMPLE 1 Find the axis of symmetry and the vertex SOLUTION For the function y = –2x2 + 12x – 7, a = 2 and b = 12. a. 12 2(– 2) x = – b 2a = = 3 Substitute – 2 for a and 12 for b. Then simplify. b. The x-coordinate of the vertex is , or 3. b 2a – y = –2(3)2 + 12(3) – 7 = 11 Substitute 3 for x. Then simplify. ANSWER The vertex is (3, 11).

EXAMPLE 2 Graph y = ax2 + b x + c Graph y = 3x2 – 6x + 2. STEP 1 Determine whether the parabola opens up or down. Because a > 0, the parabola opens up. STEP 2 = Find and draw the axis of symmetry: x = – b 2a – – 6 2(3) 1. STEP 3 Find and plot the vertex.

EXAMPLE 2 Graph y = ax2 + b x + c The x-coordinate of the vertex is – b 2a , or 1. To find the y-coordinate, substitute 1 for x in the function and simplify. y = 3(1)2 – 6(1) + 2 = – 1 So, the vertex is (1, – 1). STEP 4 Plot two points. Choose two x-values less than the x-coordinate of the vertex. Then find the corresponding y-values.

x y EXAMPLE 2 Standardized Test Practice – 1 2 11 STEP 5
– 1 y 2 11 STEP 5 Reflect the points plotted in Step 4 in the axis of symmetry. STEP 6 Draw a parabola through the plotted points.

EXAMPLE 2 Graph y = ax2 + b x + c

1. Find the axis of symmetry and vertex of the graph of the function.
GUIDED PRACTICE for Examples 1 and 2 1. Find the axis of symmetry and vertex of the graph of the function. y = x2 – 2x – 3. SOLUTION For the function y = x2 – 2x – 3, a = 1 and b = – 2. – x = – b 2a = = 1 – 2 2 1 Substitute – 2 for b and 1 for a. Then simplify. The x-coordinate of the vertex is , or 1. b 2a –

GUIDED PRACTICE for Examples 1 and 2
To find the y-coordinate, substitute 1 for x in the function and simplify. y = (1)2 – 2(1) – 3 = – 4 Substitute 1 for x. Then simplify. The vertex is (1, – 4).

GUIDED PRACTICE for Examples 1 and 2 2. Graph the Function. y = 3x2 + 12x – 1. Label the vertex and axis of symmetry. STEP 1 Determine whether the parabola opens up or down. Because a > 0, the parabola opens up. STEP 2 Find and draw the axis of symmetry: = x = – b 2a – –12 2(3) – 2. STEP 3 Find and plot the vertex.

GUIDED PRACTICE for Examples 1 and 2 The x-coordinate of the vertex is – b 2a , or – 2. To find the y-coordinate, substitute –2 for x in the function and simplify. y = 3(– 2)2 + 12(– 2) – 1 = – 14 So, the vertex is (– 2, – 14).

x y GUIDED PRACTICE for Examples 1 and 2 STEP 4 – 1
Plot two points. Choose two x-values less than the x-coordinate of the vertex. Then find the corresponding y-values. x – 1 y – 10

GUIDED PRACTICE for Examples 1 and 2 STEP 5 Reflect the points plotted in Step 4 in the axis of symmetry. STEP 6 Draw a parabola through the plotted points.

Find the minimum or maximum value
EXAMPLE 3 Find the minimum or maximum value Tell whether the function f(x) = – 3x2 – 12x + 10 has a minimum value or a maximum value. Then find the minimum or maximum value. SOLUTION Because a = – 3 and – 3 < 0, the parabola opens down and the function has a maximum value. To find the maximum value, find the vertex. x = – = – = – 2 b 2a – 12 2(– 3) The x-coordinate is – b 2a f(– 2) = – 3(– 2)2 – 12(– 2) + 10 = 22 Substitute 22 for x. Then simplify.

EXAMPLE 3 Find the minimum or maximum value ANSWER The maximum value of the function is f(– 2) = 22.

EXAMPLE 4 Find the minimum value of a function Suspension Bridges The suspension cables between the two towers of the Mackinac Bridge in Michigan form a parabola that can be modeled by the graph of y = x2 – 0.37x where x and y are measured in feet.What is the height of the cable above the water at its lowest point?

Find the minimum value of a function
EXAMPLE 4 Find the minimum value of a function SOLUTION The lowest point of the cable is at the vertex of the parabola. Find the x-coordinate of the vertex. Use a = and b = – 0.37. x = – = – = 1910 b 2a – 0.37 2( ) Use a calculator. Substitute 1910 for x in the equation to find the y-coordinate of the vertex. ≈ (1910)2 – 0.37(1910) ≈ 196

EXAMPLE 4 Find the minimum value of a function ANSWER The cable is about 196 feet above the water at its lowest point.

GUIDED PRACTICE for Examples 3 and 4 3. Tell whether the function f(x) = 6x2 + 18x + 13 has a minimum value or a maximum value. Then find the minimum or maximum value. SOLUTION Because a = 6 and 6 > 0, the parabola opens down and the function has a minimum value. To find the minimum value, find the vertex.  1 2 The minimum value is

GUIDED PRACTICE for Examples 3 and 4 Suspension Bridges 4. The cables between the two towers of the Takoma Narrow Bridge form a parabola that can be modeled by the graph of the equation y = x2 – 0.4x where x and y are measured in feet.What is the height of the cable above the water at its lowest point? Round your answer to the nearest foot.

The lowest point of the cable is at the vertex of the
GUIDED PRACTICE for Examples 3 and 4 SOLUTION The lowest point of the cable is at the vertex of the parabola. Find the x-coordinate. Use a = and b = – 0.4. x = – = – = 1428 b 2a – 0.4 2( ) Use a calculator. Substitute 1428 for x in the equation to find the y-coordinate of the vertex. y = (1428)2 – 0.4(1428) = 221

GUIDED PRACTICE for Examples 3 and 4 ANSWER The cable is about 221 feet above the water at its lowest point.

Warm-Up – 10.3 – Solve Quad. Eqns by Graphing

Find the zeros of the polynomial function BY FACTORING! 1. f(x) = x2 + 5x – 36 ANSWER –9, 4 2. f(x) = x2 – 9x + 20 ANSWER 4, 5

Find the line of symmetry and the vertex of the following: 1. f(x) = x2 + 5x – 36 ANSWER Line of symmetry  x = -2.5 Vertex = (-2.5,-42.3) Find the line of symmetry and the minimum of the following: 2. f(x) = x2 – 9x + 20 ANSWER Line of symmetry  x = 4.5 Vertex = (4.5,-0.25)

Solve. You are making a rectangular flag for your school team. The number of square inches in the area of the flag is 9x The number of inches in the length of the flag is x What is the width of the flag? HINT: DRAW A PICTURE ON YOUR WHITEBOARD! 3. ANSWER 9 in.

Find the zeros of the polynomial function by graphing. 1. f(x) = x2 + 5x – 36 ANSWER –9, 4 BONUS: Find the average of the x-coordinates of the roots. Look familiar? Bet it only works once!  2. f(x) = x2 – 9x + 20 BONUS: Find the average of the x-coordinates of the roots. Look familiar? Did it work again?!?? ANSWER 4, 5

Vocabulary – 10.3 – Quadratic Equation
An equation that can be written in the form f(x) = ax2 + bx + c roots Where an equation crosses the X-axis. AKA the ZEROS of a function. Found by setting an equation = 0 and solving for x.

Notes – 10.3–Solve Quad. Eqns by Graphing.
FINDING ZEROS OF A QUAD. FUNCTION Factoring Checking a table Graphing and finding the “zero” Quadratic Equations can have a MAXIMUM of _____ roots, but it can have _____ or even _______. Most of the time, the zeros of a function are NOT integers! 1 2 NONE

Examples 10.3

Solve a quadratic equation having two solutions
EXAMPLE 1 Solve a quadratic equation having two solutions Solve x2 – 2x = 3 by graphing. SOLUTION STEP 1 Write the equation in standard form. x2 – 2x = 3 Write original equation. x2 – 2x – 3 = 0 Subtract 3 from each side. STEP 2 Graph the function y = x2 – 2x – 3. The x-intercepts are – 1 and 3.

Solve a quadratic equation having two solutions
EXAMPLE 1 Solve a quadratic equation having two solutions ANSWER The solutions of the equation x2 – 2x = 3 are – 1 and 3. CHECK: You can check – 1 and 3 in the original equation. x2 – 2x = 3 x2 – 2x = 3 Write original equation. (–1)2 –2(–1) 3 = ? (3)2 –2(3) 3 = ? Substitute for x. 3 = 3 3 = 3 Simplify. Each solution checks.

Solve a quadratic equation having one solution
EXAMPLE 2 Solve a quadratic equation having one solution Solve – x2 + 2x = 1 by graphing. SOLUTION STEP 1 Write the equation in standard form. – x2 + 2x = 1 Write original equation. – x2 + 2x – 1 = 0 Subtract 1 from each side.

EXAMPLE 2 Solve a quadratic equation having one solution STEP 2 Graph the function y = – x2 + 2x – 1. The x-intercept is 1. ANSWER The solution of the equation – x2 + 2x = 1 is 1.

Solve a quadratic equation having no solution
EXAMPLE 3 Solve a quadratic equation having no solution Solve x2 + 7 = 4x by graphing. SOLUTION STEP 1 Write the equation in standard form. x = 4x Write original equation. x2 – 4x + 7 = 0 Subtract 4x from each side.

EXAMPLE 3 Solve a quadratic equation having no solution STEP 2 Graph the function y = x2 – 4x + 7. ANSWER The equation x2 + 7 = 4x has no solution.

GUIDED PRACTICE for Examples 1, 2 and 3 Solve the equation by graphing. 1. x2 – 6x + 8 = 0 Graph the function y = x2 – 6x + 8 The x-intercept 2 and 4. ANSWER The solution of the equation x2 – 6x + 8 are 2 and 4.

You can check – 1 and 3 in the original equation.
GUIDED PRACTICE for Examples 1, 2 and 3 CHECK: You can check – 1 and 3 in the original equation. x2 – 6x + 8 = 0 x2 – 6x + 8 = 0 Write original equation. (2)2 – = 0 x2 – 6(4) + 8 = 0 Substitute for x. 0 = 0 0 = 0 Simplify.Solution checks.

Write the equation in standard form. x2 + x = – 1 x2 + x + 1 = 0
GUIDED PRACTICE for Examples 1, 2 and 3 2. x2 + x = – 1 SOLUTION STEP 1 Write the equation in standard form. x2 + x = – 1 Write original equation. x2 + x + 1 = 0 Subtract – 1 from each side. STEP 2 Graph the function y = x2 + x The graph has no x-intercepts. The equation x2 + x = – 1 has no solution.

Write the equation in standard form. – x2 + 6x = 9
GUIDED PRACTICE for Examples 1, 2 and 3 3. – x2 + 6x = 9 SOLUTION STEP 1 Write the equation in standard form. – x2 + 6x = 9 Write original equation. – x2 + 6x – 9 = 0 Subtract 9 from each side. STEP 2 Graph the function y = – x2 + 6x – 9 The x-intercepts is 3. ANSWER The solution of the equation – x2 + 6x – 9 is 3.

EXAMPLE 6 Solve a multi-step problem Sports An athlete throws a shot put with an initial vertical velocity of 40 feet per second as shown. a. Write an equation that models the height h (in feet) of the shot put as a function of the time t (in seconds) after it is thrown. b. Use the equation to find the time that the shot put is in the air.

Solve a multi-step problem
EXAMPLE 6 Solve a multi-step problem a. Use the initial vertical velocity and the release height to write a vertical motion model. h = – 16t2 + vt + s Vertical motion model h = – 16t2 + 40t + 6.5 Substitute 40 for v and 6.5 for s. b. The shot put lands when h = 0. To find the time t when h = 0, solve 0 = – 16t2 + 40t for t. To solve the equation, graph the related function h = – 16t2 + 40t on a graphing calculator. Use the trace feature to find the t-intercepts.

EXAMPLE 6 Solve a multi-step problem ANSWER There is only one positive t-intercept. The shot put is in the air for about 2.6 seconds.

GUIDED PRACTICE for Example 6 WHAT IF? In Example 6, suppose the initial vertical velocity is 30 feet per second. Find the time that the shot put is in the air. 6.

a. Use the initial vertical velocity and the release height
GUIDED PRACTICE for Example 6 a. Use the initial vertical velocity and the release height to write a vertical motion model. h = – 16t2 + vt + s Vertical motion model h = – 16t2 + 30t + 6.5 Substitute 40 for v and 6.5 for s. b. The shot put lands when h = 0. To find the time t when h = 0, solve 0 = – 16t2 + 30t for t. ANSWER There is only one positive t-intercept. The shot put is in the air for about 2.1 seconds.

Warm-Up – 10.4

Solve the equation 1. x= – 49 √ ANSWER X = –7 2. x2 = 200 ANSWER X ≈14.1 and x≈-14.1

x = ±3 Lesson 10.4, For use with pages 652-658
Solve the following equation. 3. 4x2 = 36 ANSWER x = ±3 A ball is dropped from a height 9 feet above the ground. How long does it take the ball to hit the ground? 4. ANSWER 0.75 sec

Vocabulary – 10.4 Square Root
What number multiplied by itself gives the original If a2 = b, then b is the square root of a. Perfect Square Integers that have integer square roots.

Notes – 10.4– Use Square Roots to solve Quad. Eqns.
FINDING ROOTS OF QUAD. EQNS Factoring (GCM and binomials) Tables Graphing “What’s the goal of solvign every alg. Eqn. you will ever see??? What process did we use to do this?? If the quadratic equations are “simple” (i.e. no “bx” term), the easiest solution may be SSADMEP.

Examples 10.4

Solve quadratic equations
EXAMPLE 1 Solve quadratic equations Solve the equation. a. 2x2 = 8 SOLUTION a. 2x2 = 8 Write original equation. x2 = 4 Divide each side by 2. x = ± 4 = ± 2  Take square roots of each side. Simplify. The solutions are – 2 and 2. ANSWER

EXAMPLE 1 Solve quadratic equations b. m2 – 18 = – 18 Write original equation. m2 = 0 Add 18 to each side.. m = 0 The square root of 0 is 0. ANSWER The solution is 0.

EXAMPLE 1 Solve quadratic equations c. b = 5 Write original equation. b2 = – 7 Subtract 12 from each side. ANSWER Negative real numbers do not have real square roots. So, there is no solution.

Take square roots of a fraction
EXAMPLE 2 Take square roots of a fraction Solve 4z2 = 9. SOLUTION 4z2 = 9. Write original equation. z2 = 9 4 Divide each side by 4. z = ±  9 4 Take square roots of each side. z = ± 3 2 Simplify.

EXAMPLE 2 Take square roots of a fraction ANSWER The solutions are – and 3 2

Approximate solutions of a quadratic equation
EXAMPLE 3 Approximate solutions of a quadratic equation Solve 3x2 – 11 = 7. Round the solutions to the nearest hundredth. SOLUTION 3x2 – 11 = 7 Write original equation. 3x2 = 18 Add 11 to each side. x2 = 6 Divide each side by 3. x = ± 6  Take square roots of each side.

Approximate solutions of a quadratic equation
EXAMPLE 3 Approximate solutions of a quadratic equation x ± 2.45 Use a calculator. Round to the nearest hundredth. ANSWER The solutions are about – 2.45 and about 2.45.

Solve quadratic equations for Examples 1,2 and 3
GUIDED PRACTICE Solve quadratic equations for Examples 1,2 and 3 Solve the equation. 1. c2 – 25 = 0 SOLUTION c2 – 25 = 0 Write original equation. c = ± 25 = ± 5  Take square roots of each side. Simplify. ANSWER The solutions are – 5 and 5.

GUIDED PRACTICE Solve quadratic equations for Examples 1,2 and 3 Solve the equation. w = – 8 SOLUTION 5w = – 8 Write original equation. 5w2 = – 8 –12 Subtract 12 from each side.  w = –4 Take square roots of each side. Simplify. ANSWER Negative real numbers do not have a real square root. So there is no solution.

GUIDED PRACTICE Solve quadratic equations for Examples 1,2 and 3 Solve the equation. x = 11 SOLUTION 2x = 11 Write original equation. 2x2 = 0 Subtract 11 from each side. x = 0 The root of 0 is 0. ANSWER The solution is 0 .

 EXAMPLE 1 GUIDED PRACTICE Solve quadratic equations
for Examples 1,2 and 3 Solve the equation. x2 = 16 SOLUTION 25x2 = 16 Write original equation. x = 16 25 Divided each to by 25. x = ±  16 25 Take square roots of each side. x = ± 4 5 Simplify. ANSWER The solution is – and 4 5

for Examples 1,2 and 3 Solve the equation. m2 = 100 SOLUTION 9m2 = 100 Write original equation. m = 100 9 Divided each to by 9. m = ±  100 9 Take square roots of each side. m = ± 10 3 Simplify. ANSWER The solution is – and 10 3

for Examples 1,2 and 3 Solve the equation. b = 0 SOLUTION 49b = 0 Write original equation. 49b2 = – 64 Subtract 64 from each side. b2 = –64 49 Divided each to by 9. b = – 64 49  Take square roots of each side. ANSWER Negative real numbers do have real square root. So there is no solution.

for Examples 1,2 and 3 Solve the equation. Round the solution to the nearest hundredth. 7. x2 + 4 = 14 SOLUTION x2 + 4 = 14 Write original equation. x2 = 10 Subtract 4 from each side. x =  10 + – Take square roots of each side. x = + – 3.16 Use a calculation. Round to the nearest hundredth. ANSWER The solutions are about – 3.16 and 3.16.

for Examples 1,2 and 3 Solve the equation. Round the solution to the nearest hundredth. k2 – 1 = 0 SOLUTION 3k2 – 1 = 0 Write original equation. 3k2 = 1 Add 1 to each side. k2 = 1 3 Divided each to by 3. k =  + – 1 3 Take square roots of each side. k = + – 0.58 Use a calculation. Round to the nearest hundredth.

Solve a quadratic equation
EXAMPLE 4 Solve a quadratic equation Solve 6(x – 4)2 = 42. Round the solutions to the nearest hundredth. 6(x – 4)2 = 42 Write original equation. (x – 4)2 = 7 Divide each side by 6. x – 4 = ± 7  Take square roots of each side. 7  x = 4 ± Add 4 to each side. ANSWER The solutions = and – 7 

EXAMPLE 4 Solve a quadratic equation CHECK To check the solutions, first write the equation so that 0 is on one side as follows: 6(x – 4)2 – 42 = 0. Then graph the related function y = 6(x – 4)2 – 42. The x-intercepts appear to be about 6.6 and about 1.3. So, each solution checks.

for Examples 4 and 5 Solve the equation. Round the solution to the nearest hundredth. (x – 2)2 = 18 SOLUTION 2(x – 2)2 = 18 Write original equation. (x – 2)2 = 9 Divided each to by 2. x – 2 = + –  9 Take square roots of each side. x = 2 + 3 Add 2 to each side. ANSWER The solutions are about – 1 and 5.

   EXAMPLE 1 GUIDED PRACTICE Solve quadratic equations
for Examples 4 and 5 Solve the equation. Round the solution to the nearest hundredth. (q – 3)2 = 28 SOLUTION 4(q – 3)2 = 28 Write original equation. (q – 3)2 = 7 Divided each side by 4. q – 3 = + –  7 Take square roots of each side. q = 3 +  7 – Add 3 to each side. ANSWER The solutions are and = 5.65 and = 0.35 3 +  7 3 –

   EXAMPLE 1 GUIDED PRACTICE Solve quadratic equations
for Examples 4 and 5 Solve the equation. Round the solution to the nearest hundredth. (t – 5)2 = 24 SOLUTION 3(t – 5)2 = 24 Write original equation. (t – 5)2 = 8 Divided each to by 3. t – 5 = + –  8 Take square roots of each side. t =  8 – Add 5 to each side. ANSWER The solutions are and – 2.17 and = –7.83. 5 +  8 – 5 –

EXAMPLE 5 Solve a multi-step problem Sports Event During an ice hockey game, a remote-controlled blimp flies above the crowd and drops a numbered table-tennis ball. The number on the ball corresponds to a prize. Use the information in the diagram to find the amount of time that the ball is in the air.

EXAMPLE 5 Solve a multi-step problem SOLUTION STEP 1 Use the vertical motion model to write an equation for the height h (in feet) of the ball as a function of time t (in seconds).

EXAMPLE 5 Solve a multi-step problem h = – 16t2 + vt + s Vertical motion model h = – 16t2 + 0t + 45 Substitute for v and s. STEP 2 Find the amount of time the ball is in the air by substituting 17 for h and solving for t.

 EXAMPLE 5 Solve a multi-step problem h = – 16t2 + 45
Write model. 17 = – 16t2 + 45 Substitute 17 for h. – 28 = – 16t2 Subtract 45 from each side. 28 16 = t2 Divide each side by 16.  28 16 = t Take positive square root. t Use a calculator.

EXAMPLE 5 Solve a multi-step problem ANSWER The ball is in the air for about 1.32 seconds

EXAMPLE 1 GUIDED PRACTICE Solve quadratic equations for Examples 4 and 5 WHAT IF? In Example 5, suppose the table-tennis ball is released 58 feet above the ground and is caught 12 feet above the ground. Find the amount of time that the ball is in the air. Round your answer to the nearest hundredth of a second. 13.

Solve quadratic equations for Examples 4 and 5
GUIDED PRACTICE Solve quadratic equations for Examples 4 and 5 SOLUTION STEP 1 Use the vertical motion model to write an equation for the height h (in feet) of the ball as a function of time t (in seconds). h = – 16t2 + vt + s Vertical motion model h = – 16t2 + 0t + 45 Substitute for v and s. STEP 2 Find the amount of time the ball is in the air by substituting 12 for h and solving for t.

for Examples 4 and 5 h = – 16t2 + 58 Write model. 12 = – 16t2 + 58 Substitute 12 for h. – 46 = – 16t2 Subtract 58 from each side. 46 16 = t2 Divide each side by – 16.  46 16 = t Take positive square root. t Use a calculator. ANSWER The ball is in the are for about 1.70 second.

Warm-Up – 10.6

1. Solve |x – 6| = 4. 1. Solve x2 + 5x = -6. ANSWER 2, 10 ANSWER x = -2 or x = -3 Solve by graphing: x2 + 5x = -5 Round to the nearest tenth. 2. Solve |x + 5| – 8 = 2. ANSWER –15, 5 ANSWER x = -3.6 or x = -1.4

Vocabulary – 10.6 Quadratic Formula
Formula that will provide the roots (real and “non-real”) for ALL quadratic equations

Notes – 10.6–Solve Quad. Eqns. With the Quadratic Formula
The Quadratic Formula below will allow you to find the roots of ALL quadratic equations. Use this when: It isn’t easily factorable. The numbers are so big or small that using a calculator is difficult b/c of the window settings. You don’t know how else to do it!

Examples 10.6

Standardized Test Practice
EXAMPLE 1 Standardized Test Practice What are the solutions of 3x2 + 5x = 8? – 1 and – A 8 3 B – 1 and 8 3 C 1 and – 8 3 D 1 and 8 3 SOLUTION 3x2 + 5x = 8 Write original equation. 3x2 + 5x – 8 = 0 Write in standard form. x = – b b2 – 4ac 2a Quadratic formula

Standardized Test Practice
EXAMPLE 1 Standardized Test Practice Substitute values in the quadratic formula: a = 3, b = 5, and c = – 8. x = – 5 52 – 4(3)(– 8) 2(3) = – 5 121 6 Simplify. = – 5 ± 11 6 Simplify the square root. The solutions of the equation are – 6 = 1 and – 5 – 11 = – 8 3 ANSWER The correct answer is C. A B D C

Solve a quadratic equation
EXAMPLE 2 Solve a quadratic equation Solve 2x2 – 7 = x. 2x2 – 2 = x Write original equation. 2x2 – x – 7 = 0 Write in standard form. x = b2 – 4ac + – –b 2a Quadratic formula Substitute values in the quadratic formula: a = 2, b = – 1, and c = – 7. – (– 1) – + ( –1)2 – 4(2)(– 7) 2(2) = 4 = + – Simplify.

EXAMPLE 2 Solve a quadratic equation ANSWER 57 1 + 2.14 and 1 – 4 – 1.64. The solutions are CHECK Write the equation in standard form, 2x2 – x – 7 = 0. Then graph the related function y = 2x2 – x – 7. The x-intercepts are about – 1.6 and 2.1. So, each solution checks.

Solve the equation by Quadratic Formula. x2 – 8x +16 = 0. 1
GUIDED PRACTICE for Examples 1and 2 Solve the equation by Quadratic Formula. x2 – 8x +16 = 0. 1 SOLUTION x2 – 8x +16 = 0 Write original equation. x = b2 – 4ac + – –b 2a Quadratic formula Substitute values in the quadratic formula: a = 1, b = – 8, and c = 16. – – 8 – + ( –8)2 – x= 2 1 2 = + – Simplify. ANSWER The solution of the equation is 4.

Solve the equation by Quadratic Formula. 3n2 – 5n = – 1. 2
GUIDED PRACTICE for Examples 1and 2 Solve the equation by Quadratic Formula. 3n2 – 5n = – 1. 2 SOLUTION 3n2 – 5n = – 1 Write original equation. 3n2 – 5n +1 = 0 Write in standard form. x = b2 – 4ac + – –b 2a Quadratic formula Substitute values in the quadratic formula: a = 3, b =( – 5), and c = 1. –(–5) – + ( –5)2 – x= 2 1 6 = + – Simplify.

GUIDED PRACTICE for Examples 1and 2 ANSWER The solutions are 6 = = and 5 – 3 = CHECK Write the equation in standard form, 3n2 – 5n + 1 = 0. Then graph the related function y = 3n2 –5n The x-intercepts are about 0.23 and 1.4.3s So, each solution checks.

Solve the equation by Quadratic Formula. 4z2 = +7z +2 2
GUIDED PRACTICE for Examples 1and 2 Solve the equation by Quadratic Formula. 4z2 = +7z +2 2 SOLUTION 4z2 = +7z +2 Write original equation. 4z2 –7z – 2 = 0 Write in standard form. x = b2 – 4ac + – –b 2a Quadratic formula Substitute values in the quadratic formula: a = 4, b = 7, and c = 2. –(–7) – + ( –7)2 – (2) x= 2 4 8 = + – Simplify.

GUIDED PRACTICE for Examples 1and 2 ANSWER The Solution are 8 = = 2.0 and 7 – = 0.25

Use the quadratic formula
EXAMPLE 3 Use the quadratic formula Film Production For the period , the number y of films produced in the world can be modeled by the function y = 10x x where x is the number of years since In what year were 4200 films produced? – SOLUTION y = 10x2 – 94x Write function. 4200 = 10x2 – 94x – 3900 Substitute 4200 for y. 0 = 10x2 – 94x – 300 Write in standard form.

Use the quadratic formula
EXAMPLE 3 Use the quadratic formula x = (–94)2 – 4 (10)(–300) –(–94) + – 2(10) Substitute values in the quadratic formula: a = 10, b = –94, and c = – 300. 20,836 94 + – 20 = Simplify. The solutions of the equation are 94 + 20,836 20 12 and 94 – 20,836 20 –3. ANSWER There were 4200 films produced about 12 years after 1971, or in 1983.

WHAT IF? In Example 3, find the year when 4750 films were produced.
GUIDED PRACTICE for Example 4 4. WHAT IF? In Example 3, find the year when 4750 films were produced. SOLUTION y = 10x2 – 94x Write function. 4750 = 10x2 – 94x Substitute 4750 for y. 0 = 10x2 – 94x – 850 Write in standard form.

The solutions of the equation are
GUIDED PRACTICE for Example 4 x = (–94)2 – 4 (10)(–850) –(–94) + – 2(10) Substitute values in the quadratic formula: a = 10, b = –94, and c = – 850. 8836 42,836 94 + – 20 = Simplify. The solutions of the equation are 94 + 42,836 20 15 and 94 – 42,836 20 –6. ANSWER There were 4750 films produced about 15 years after 1971, or in 1986.

EXAMPLE 4 Choose a solution method Tell what method you would use to solve the quadratic equation. Explain your choice(s). a. 10x2 – 7 = 0 SOLUTION a. The quadratic equation can be solved using square roots because the equation can be written in the form x2 = d.

EXAMPLE 4 Choose a solution method Tell what method you would use to solve the quadratic equation. Explain your choice(s). b. x2 + 4x = 0 SOLUTION b. The equation can be solved by factoring because the expression x2 + 4x can be factored easily. Also, the equation can be solved by completing the square because the equation is of the form ax2 + bx + c = 0 where a = 1 and b is an even number.

EXAMPLE 4 Choose a solution method Tell what method you would use to solve the quadratic equation. Explain your choice(s). c. 5 x2 + 9x – 4 = 0 SOLUTION c. The quadratic equation cannot be factored easily, and completing the square will result in many fractions. So, the equation can be solved using the quadratic formula.

GUIDED PRACTICE for Example 4 Tell what method you would use to solve the quadratic equation. Explain your choice(s). x2 +x – 6 = 0 5. SOLUTION Factoring because the expression factors easily.

GUIDED PRACTICE for Example 4 Tell what method you would use to solve the quadratic equation. Explain your choice(s). x2– 9 = 0 6. SOLUTION Factoring because the expression factors easily. Using square roots is another option since the equation can be written in the form x2 = d.

GUIDED PRACTICE for Example 4 Tell what method you would use to solve the quadratic equation. Explain your choice(s). x2 +6x =5 7. SOLUTION Completing the square because the equation is of the form ax2 + bx = c where a = 1 and b is an even number. Another method is the quadratic formula since the equation does not factor easily.

Warm-Up –

Evaluate the expression. A ball is kicked into the air from a height of 4.5 feet with an initial velocity of 30 feet per second. What is the height of the ball after 1 second? 3. ANSWER 18.5 ft

1. Graph y = 3x. ANSWER Tell whether the ordered pairs (0, 0), (1, 2), (2, 4), and (3, 6) represent a linear function, a quadratic function, or an exponential function. Why?? 2. ANSWER Linear

For y = x2 – 3x – 5, find corresponding y-values for the x-values –2, 1, and 3. 3. ANSWER 5, –7, –5

Vocabulary – 10.7-10.8 Discriminant
Value of the expression b2 – 4(a)( c) in the quadratic formula

Notes – 10.7 – Interpret the discriminant.

Notes –10.8– Linear, Quad. And Exponential func’s.
To determine which function is represented, graph the points and look at the shape, Check differences/patterns or Use the Regression functions on “Calli.”

Examples

EXAMPLE 1 Use the discriminant Equation ax2 + bx + c = 0 Discriminant b2 – 4ac Number of solutions a. 2x2 + 6x + 5 = 0 62 – 4(2)(5) = –4 No solution b. x2 – 7 = 0 02 – 4(1)(– 7) = 28 Two solutions c. 4x2 – 12x + 9 = 0 (–12)2 –4(4)(9) = 0 One solution

Find the number of solutions
EXAMPLE 2 Find the number of solutions Tell whether the equation 3x2 – 7 = 2x has two solutions, one solution, or no solution. SOLUTION STEP 1 Write the equation in standard form. 3x2 – 7 = 2x Write equation. 3x2 – 2x – 7 = 0 Subtract 2x from each side.

Find the number of solutions
EXAMPLE 2 Find the number of solutions STEP 2 Find the value of the discriminant. b2 – 4ac = (–2)2 – 4(3)(–7) Substitute 3 for a, – 2 for b, and –7 for c. = 88 Simplify. The discriminant is positive, so the equation has two solutions. ANSWER

Write the equation in standard form.
GUIDED PRACTICE for Examples 1 and 2 Tell whether the equation has two solutions, one solution, or no solution. 1. x2 + 4x + 3 = 0 SOLUTION STEP 1 Write the equation in standard form. x2 + 4x + 3 = 0 Write equation.

Find the value of the discriminant.
GUIDED PRACTICE for Examples 1 and 2 STEP 2 Find the value of the discriminant. b2 – 4ac = (– 4)2 – 4(1)(3) Substitute 1 for a, 4 for b, and 3 for c. = 4 Simplify. The discriminant is positive, so the equation has two solutions. ANSWER

EXAMPLE 4 Solve a multi-step problem Fountains The Centennial Fountain in Chicago shoots a water arc that can be modeled by the graph of the equation y=  0.006x x where x is the horizontal distance (in feet) from the river’s north shore and y is the height (in feet) above the river. Does the water arc reach a height of 50 feet? If so, about how far from the north shore is the water arc 50 feet above the water?

EXAMPLE 4 Solve a multi-step problem SOLUTION STEP 1 Write a quadratic equation. You want to know whether the water arc reaches a height of 50 feet, so let y = 50. Then write the quadratic equation in standard form. y x x 10 = – + Write given equation. x x 10 = – + Substitute 50 for y. x x 40 = – + Subtract 50 from each side.

EXAMPLE 4 Solve a multi-step problem STEP 2 Find the value of the discriminant of 0 = 0.006x x – 40. b2 4ac = (1.2)2 4(– 0.006)( 40) – = 0.48 a = – 0.006, b = 1.2, c = – 40 Simplify STEP 3 Interpret the discriminant. Because the discriminant is positive, the equation has two solutions. So, the water arc reaches a height of 50 feet at two points on the water arc.

  EXAMPLE 4 Solve a multi-step problem STEP 4
Solve the equation to find the distance from the north shore where the water arc is 50 feet above the water. 0 = 0.006x x – 40. – x =  b ac b 2a Quadratic formula = –  1.2 0.48 2( ) Substitute values in the quadratic formula. 0 x or x Use a calculator.

EXAMPLE 4 Solve a multi-step problem The water arc is 50 feet above the water about 42 feet from the north shore and about 158 feet from the north shore. ANSWER

EXAMPLE 2 Identify functions using differences or ratios Use differences or ratios to tell whether the table of values represents a linear function, an exponential function, or a quadratic function. x –2 –1 1 2 y –6 –4 6 First differences: Second differences: ANSWER The table of values represents a quadratic function.

EXAMPLE 2 Identify functions using differences or ratios x – 2 – 1 1 2 y 4 7 10 Differences: ANSWER The table of values represents a linear function.

GUIDED PRACTICE for Examples 1 and 2 1. Tell whether the ordered pairs represent a linear function, an exponential function, or a quadratic function: (0, – 1.5), (1, –0.5),(2,2.5),(3,7.5). ANSWER quadratic function

GUIDED PRACTICE for Examples 1 and 2 2. Tell whether the table of values represents a linear function, an exponential function, or a quadratic function and find the function! y 2 x – 2 – 1 1 0.08 0.4 10 *5 *5 *5 ANSWER exponential function Y = 2 * 5x

EXAMPLE 3 Write an equation for a function Tell whether the table of values represents a linear function, an exponential function, or a quadratic function. Then write an equation for the function. x – 2 – 1 1 2 y 0.5

EXAMPLE 3 Write an equation for a function SOLUTION STEP 1 Determine which type of function the table of values represents. x – 2 – 1 1 2 y 0.5 First differences: – – Second differences:

Write an equation for a function
EXAMPLE 3 Write an equation for a function The table of values represents a quadratic function because the second differences are equal. STEP 2 Write an equation for the quadratic function. The equation has the form y = ax2. Find the value of a by using the coordinates of a point that lies on the graph, such as (1, 0.5). y = ax2 Write equation for quadratic function. 0.5 = a(1)2 Substitute 1 for x and 0.5 for y. 0.5 = a Solve for a.

EXAMPLE 3 Write an equation for a function ANSWER The equation is y = 0.5x2.

GUIDED PRACTICE EXAMPLE 3 for Example 3 Tell whether the table of values represents a linear function, an exponential function, or a quadratic function. Then write an equation for the function. x – 3 – 2 – 1 1 y –7 –5 3. SOLUTION STEP 1 Determine which type of function the table of values represents.

GUIDED PRACTICE EXAMPLE 3 for Example 3 x – 3 – 2 – 1 1 y –7 –5 – 2 – 2 – 2 – 2 First difference The table of value represent a linear function.

EXAMPLE 3 GUIDED PRACTICE for Example 3 STEP 2
Write an equation for the quadratic function. The equation has the form y = ax2. Find the value of a by using the coordinates of a point that lies on the graph, such as (1, 1). y = ax2 Write equation for quadratic function. 1= a(1)2 Substitute 1 for x and 0.5 for y. 1 = a Solve for a. The equation is y = 2x - 1

GUIDED PRACTICE EXAMPLE 3 for Example 3 Tell whether the table of values represents a linear function, an exponential function, or a quadratic function. Then write an equation for the function. x – 2 – 1 1 2 y 8 4 4.

GUIDED PRACTICE EXAMPLE 3 for Example 3 SOLUTION Determine which type of function the table of values represents. x – 2 – 1 1 2 y 8 The table of values represents a quadratic function. The equation is y = 2x2

Review – Ch. 10 – PUT HW QUIZZES HERE

Daily Homework Quiz For use after Lesson 10.1 1. Graph y = –0.5x2 + 2 ANSWER 2. How would the graph of the function y = – 2x2 + 3 be affected if the function were changed to y = – 2x2 – 3? It would be shifted down 6 units. ANSWER

Daily Homework Quiz For use after Lesson 10.1 3. A pinecone falls about 50 feet from the branch of a pine tree.Its height (in feet) can be modeled by the function h(t) = -16t2 + 50, where t is the time in seconds. How long does it take to land on the ground? ANSWER about 1.8 sec

Daily Homework Quiz For use after Lesson 10.2 1. Find the axis of symmetry and the vertex of the graph of the function y = – 4x2 + 8x – 9 ANSWER x = 1; (1, – 5) 2. Graph y = –2x2 + 4x + 1 ANSWER

Daily Homework Quiz For use after Lesson 10.2 An arch to the entrance of the library can be modeled by y = – 0.13x x, where x and y are measured in feet. To the nearest foot, what is the height of the highest point of the arch? ANSWER 12 ft

Daily Homework Quiz For use after Lesson 10.3 1. Solve x2 +6x + 8 by graphing. – 4 – 2 ANSWER

Daily Homework Quiz For use after Lesson 10.3 Find the number of solutions for each equation. 2. x2 + 6x = – 10 ANSWER none 3. x2 + 6x = – 9 one ANSWER

Daily Homework Quiz For use after Lesson 10.3 4. Find the zeros of f(x) = – x2 + 2x + 3. – 1, 3 ANSWER 5. Approximate the zeros of f(x) = x2 + x – 3 to the nearest tenth. ANSWER – 2.3, 1.3

Daily Homework Quiz For use after Lesson 10.3 6. Maria throws a shot put with an initial velocity of 25 feet per second. She releases it at a height of 5 feet. Find the time the shot put is in the air. about 1.7 sec ANSWER

Daily Homework Quiz For use after Lesson 10.4 Solve the equation. Round solution to the nearest hundredth, if necessary. b2 – 13 = 3 ANSWER –2, 2 x = 25 ANSWER – 5 3 , n2 –18 = 12 ANSWER –3.16, 3.16

Daily Homework Quiz For use after Lesson 10.4 At a football game you are sitting 32 feet above the ground. If your hat comes off falls to the ground, how long will it be in the air? 4. ANSWER About 1.41 sec

Daily Homework Quiz For use after Lesson 10.5 Solve the equation by completing the square. Round to the nearest hundredth, if necessary. 1. x2 + 12x = 28 ANSWER –14, 2 2. m2 – 8m = 12 ANSWER – 1.29, 9.29

Daily Homework Quiz For use after Lesson 10.5 What is the width of the border that surrounds this poster? 3. ANSWER 1 in

Daily Homework Quiz For use after Lesson 10.6 Use the quadratic formula to solve the equation.Round your solutions to the nearest hundredth, if necessary. 1. 6x2 – 6 = – 5x , ANSWER 3 2 – 2. 2x2 + 3x – 8 = 0 –2.87, 1.39 ANSWER

Daily Homework Quiz For use after Lesson 10.6 m2 – m + 9 = 12 ANSWER – 0.85, 1.18 You have seen that the function y = 10x2 + 94x models the number y of films produced in the world, where x is the number of years since In what year were 10,000 films produced? 4. ANSWER about 29.8 years after 1971, or during the year 2000

Daily Homework Quiz For use after Lesson 10.7 Tell whether the equation has two solutions, one solution, or no solutions. 1. 4b2 + 2b – 5 = 0 2. 2g2 + 8g = – 11

Daily Homework Quiz For use after Lesson 10.7 Find the number of x-intercepts of the graphs of the equations 3. y = x2 + 14x + 49 y = x2 + 14x + 50 4.

Daily Homework Quiz For use after Lesson 10.7 5. The graph of y = – 0.2x x models the height of one of the arches at the entrance to a parking structure. Can a truck that is 20 feet high fit under the arch?

Daily Homework Quiz For Section 10.8 6. Tell whether the table of values represents a linear function, an exponential function, or a quadratic function and find the function! 1 y x – 1 2 16 4 1/4

Answers 1) ANSWER two solutions 2) ANSWER no solutions 3) ANSWER one 4) ANSWER none The value of the discriminate of 0 = – 0.2x x – 20 is negative, so there is no solution.The truck will not fit under the arch. 5) ANSWER 6) ANSWER exponential function Y = 4(1/4)x

Warm-Up – X.X

Vocabulary – X.X Holder Holder 2 Holder 3 Holder 4

Notes – X.X – LESSON TITLE.
Holder

Examples X.X

Unit 6 – Chapter 10.

Similar presentations

Presentation on theme: "Unit 6 – Chapter 10."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Unit 6 – Chapter 10.

Similar presentations

Presentation on theme: "Unit 6 – Chapter 10."— Presentation transcript:

Similar presentations

About project

Feedback