 # LIAL HORNSBY SCHNEIDER

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LIAL HORNSBY SCHNEIDER
COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER

Graphing Techniques Completing the Square The Vertex Formula Quadratic Models and Curve Fitting

Polynomial Function A polynomial function of degree n, where n is a nonnegative integer, is a function defined by an expression of the form where an, an-1, …, a1, and a0 are real numbers, with an ≠ 0.

Polynomial Function For the polynomial function defined by
n is 3 and the polynomial has the form

where a, b, and c are real numbers, with a ≠ 0.

Simplest Quadratic x (x) – 2 4 – 1 1 2 range [0, ) y x
3 – 2 4 – 3 – 4 x (x) – 2 4 – 1 1 2 x domain (−, )

Simplest Quadratic Parabolas are symmetric with respect to a line. The line of symmetry is called the axis of the parabola. The point where the axis intersects the parabola is the vertex of the parabola. Opens up Vertex Axis Axis Vertex Opens down

Applying Graphing Techniques to a Quadratic Function
The graph of g(x) = ax2 is a parabola with vertex at the origin that opens up if a is positive and down if a is negative. The width of the graph of g(x) is determined by the magnitude of a. The graph of g(x) is narrower than that of (x) = x2 if a> 1 and is broader (wider) than that of (x) = x2 if a< 1. By completing the square, any quadratic function can be written in the form the graph of F(x) is the same as the graph of g(x) = ax2 translated hunits horizontally (to the right if h is positive and to the left if h is negative) and translated k units vertically (up if k is positive and down if k is negative).

Graph the function. Give the domain and range.
GRAPHING QUADRATIC FUNCTIONS Example 1 Graph the function. Give the domain and range. a. 2 3 – 2 – 6 x (x) – 1 3 – 2 1 – 5 2 – 6 4 5 Solution Domain (−, ) Range [– 6, )

Graph the function. Give the domain and range.
GRAPHING QUADRATIC FUNCTIONS Example 1 Graph the function. Give the domain and range. 2 3 – 2 – 6 b. Solution Domain (−, ) Range (–, 0]

Graph the function. Give the domain and range.
GRAPHING QUADRATIC FUNCTIONS Example 1 Graph the function. Give the domain and range. (4, 3) 3 – 2 – 6 x = 4 c. Solution Domain (−, ) Range (–, 3]

Graph by completing the square and locating the vertex.
GRAPHING A PARABOLA BY COMPLETING THE SQUARE Example 2 Graph by completing the square and locating the vertex. Solution Express x2– 6x + 7 in the form (x– h)2 + k by completing the square. Complete the square.

Graph by completing the square and locating the vertex.
GRAPHING A PARABOLA BY COMPLETING THE SQUARE Example 2 Graph by completing the square and locating the vertex. Solution Express x2 – 6x + 7 in the form (x– h)2 + k by completing the square. Add and subtract 9. Regroup terms. Factor; simplify. This form shows that the vertex is (3, – 2)

Graph by completing the square and locating the vertex.
GRAPHING A PARABOLA BY COMPLETING THE SQUARE Example 2 Graph by completing the square and locating the vertex. Solution Find additional ordered pairs that satisfy the equation. Use symmetry about the axis of the parabola to find other ordered pairs. Connect to obtain the graph. Domain is (−, ) Range is [–2, )

Motion Problems Note In Example 2 we added and subtracted 9 on the same side of the equation to complete the square in Section This differs from adding the same number to each side of the equation, as when we completed the square. Since we want (x) (or y) alone on one side of the equation, we adjusted that step in the process of completing the square.

Graph by completing the square and locating the vertex.
GRAPHING A PARABOLA BY COMPLETING THE SQUARE Example 3 Graph by completing the square and locating the vertex. Solution To complete the square, the coefficient of x2 must be 1. Factor – 3 from the first two terms.

Graph by completing the square and locating the vertex.
GRAPHING A PARABOLA BY COMPLETING THE SQUARE Example 3 Graph by completing the square and locating the vertex. Solution Distributive property Be careful here. Factor; simplify.

Graph by completing the square and locating the vertex.
GRAPHING A PARABOLA BY COMPLETING THE SQUARE Example 3 Graph by completing the square and locating the vertex. Solution Factor; simplify.

Graph by completing the square and locating the vertex.
GRAPHING A PARABOLA BY COMPLETING THE SQUARE Example 3 Graph by completing the square and locating the vertex. Solution Intercepts are good additional points to find. Here is the y-intercept. Let x = 0.

Graph by completing the square and locating the vertex.
GRAPHING A PARABOLA BY COMPLETING THE SQUARE Example 3 Graph by completing the square and locating the vertex. Solution The x-intercepts are found by setting (x) equal to 0 in the original equation. Let (x) = 0. Multiply by –1; rewrite. Factor. Zero-factor property

GRAPHING A PARABOLA BY COMPLETING THE SQUARE Example 3
2 The y-intercept is 1 This x-intercept is 1/3. This x-intercept is – 1.

The quadratic function defined by (x) = ax2 + bx + c can be written as where

The graph of  has the following characteristics. It is a parabola with vertex (h, k) and the vertical line x = h as axis. It opens up if a > 0 and down is a < 0. It is broader than the graph of y = x2 if a< 1 and narrower if a> 1. The y-intercept is (0) = c. If b2 – 4ac > 0, the x-intercepts are If b2 – 4ac = 0, the x-intercepts is If b2 – 4ac < 0, there are no x-intercepts.

Since (– 1) = 2(– 1)2 + 4 (– 1) +5 = 3, the vertex is (– 1, 3).
FINDING THE AXIS AND THE VERTEX OF A PARABOLA USING THE VERTEX FORMULA Example 4 Find the axis and vertex of the parabola having equation (x) = 2x2 +4x + 5 using the vertex formula. Solution Here a = 2, b = 4, and c = 5. The axis of the parabola is the vertical line The vertex is (– 1, (– 1)). Since (– 1) = 2(– 1)2 + 4 (– 1) +5 = 3, the vertex is (– 1, 3).

Quadratic functions make good models for data sets where the data either increases, levels off, and then decreases, levels off, and then increases. An application that models the path of the projectile is

Solution Use the projectile height function with v0 = 80 and s0 = 100.
SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Example 5 A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. a. Give the function that describes the height of the ball in terms of time t. Solution Use the projectile height function with v0 = 80 and s0 = 100.

Solution One choice for a window is [-.3, 9.7] by [-60,300]
SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Example 5 A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. b. Graph this function on a graphing calculator so that the y-intercept, the positive x-intercept, and the vertex are visible. Solution One choice for a window is [-.3, 9.7] by [-60,300]

SOLVING A PROBLEM INVOLVING PROJECTILE MOTION
Example 5 A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. c. The point (4.8, ) lies on the graph of the function. What does this mean for this particular situation? Solution When 4.8 seconds have elapsed, the projectile is at a height of ft.

SOLVING A PROBLEM INVOLVING PROJECTILE MOTION
Example 5 A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. d. After how many seconds does the projectile reach its maximum height? What is this maximum height? Solution Find the coordinates of the vertex of the parabola. a = – 16 and b = 80

After 2.5 sec the ball reaches its maximum height of 200 ft.
SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Example 5 A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. d. After how many seconds does the projectile reach its maximum height? What is this maximum height? Solution Find the coordinates of the vertex of the parabola. a = – 16 and b = 80 and After 2.5 sec the ball reaches its maximum height of 200 ft.

SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Example 5 A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. e. For what interval of time is the height of the ball greater than 160 ft? Solution Solve the quadratic inequality. Subtract 160. Divide by – 4; reverse the inequality symbol.

Solution By the quadratic formula, the solutions are…
SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Example 5 A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. e. For what interval of time is the height of the ball greater than 160 ft? Solution By the quadratic formula, the solutions are…

Solution The intervals are (– , .92), (.92, 4.08), and (4.08, ).
SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Example 5 A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. e. For what interval of time is the height of the ball greater than 160 ft? Solution The intervals are (– , .92), (.92, 4.08), and (4.08, ). A test value in each interval shows that (.92, 4.08) satisfies the inequality. The ball is more than 160 ft above ground between .92 sec and 4.08 sec.

f. After how many seconds will the ball hit the ground?
SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Example 5 A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. f. After how many seconds will the ball hit the ground? Solution The height is zero when the ball hits the ground. Find the positive solution…

f. After how many seconds will the ball hit the ground?
SOLVING A PROBLEM INVOLVING PROJECTILE MOTION Example 5 A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. f. After how many seconds will the ball hit the ground? Solution The ball hits the ground after about 6.04 sec.

MODELING THE NUMBER OF HOSPITAL OUTPATIENT VISITS
Example 6 80 represents 1980, 100 represents 2000, and so on, and the number of outpatient visits is given in millions. a. Prepare a scatter diagram, and determine a quadratic model for these data. Year Visits 80 263.0 99 573.5 90 368.2 100 592.7 95 483.2 101 612.0 96 505.5 102 640.5 97 520.6 103 648.6 98 545.5 104 662.1

MODELING THE NUMBER OF HOSPITAL OUTPATIENT VISITS
Example 6 80 represents 1980, 100 represents 2000, and so on, and the number of outpatient visits is given in millions. a. Prepare a scatter diagram, and determine a quadratic model for these data. Solution

b. Predict the number of visits in 2008.
MODELING THE NUMBER OF HOSPITAL OUTPATIENT VISITS Example 6 80 represents 1980, 100 represents 2000, and so on, and the number of outpatient visits is given in millions. b. Predict the number of visits in 2008. Solution

b. Predict the number of visits in 2008.
MODELING THE NUMBER OF HOSPITAL OUTPATIENT VISITS Example 6 80 represents 1980, 100 represents 2000, and so on, and the number of outpatient visits is given in millions. b. Predict the number of visits in 2008. Solution Since 2008 corresponds to x = 108, the model predicts that in 2008 the number of visits will be…

b. Predict the number of visits in 2008.
MODELING THE NUMBER OF HOSPITAL OUTPATIENT VISITS Example 6 b. Predict the number of visits in 2008.