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Notes: Graphing Quadratic Functions and solving quadratic linear systems algebraically Aim: Students will be able to solve quadratic linear systems graphically and algebraically. Grab your foldable that I gave you from class today and start filling in your notes. Happy “foldabling”.

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What is a quad ratic- line ar system? A ____________system consists of a __________ equation and a ________equation. The _______of a quadratic linear system is the ______________of numbers that make both equations true. Depending on how many times the line _________ the curve, the solution set may contain ____ ordered pairs, ___ ordered pair, or __ ordered pairs. quadratic linearquadratic linear solution set of ordered pairs intersects twoone no 2 solutions1 solution no solution

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Solve the quadratic-linear system graphically: 1. Draw the graph of y = x 2 – 6x + 6 2. Find the axis of symmetry for the graph using 3. Then construct the table of values for x less than 3 and x greater than 3. 4. Graph the line y – x = -4 using slope-intercept form. 5. The points where the graphs intersect are the solution to the system. E XAMPLE 1 xy 0 1 2 3 4 5 6 [(2,-2) and (5,1)] 6 1 -2 -3 -2 1 6 y = x 2 – 6x + 6 y – x = -4 y = x 2 – 6x + 6 The quadratic-linear system has ______ solution(s). The solution(s) _________________ 2 (2, -2) and (5, 1) + x y = x – 4 slope(m) = 1 y-int(b) = -4 y = x 2 – 6x + 6 Don’t forget to label the graphs. a = 1 b = -6 c = 6 Show work for line in foldable, underneath box “Graph a line using…”

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Check with graphing calculator Type the 2 functions into your graphing calculator. y 1 = x 2 + 6x + 6 AND the line y 2 = x – 4 on the interval [0,6] xy1y1 y2y2 0 6-4 1 1-3 2 -2 3 -3 4 -20 5 11 6 62 Solution to the system: {(-2, -2), (1, 1)} quadraticline

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Solve the quadratic-linear system graphically: 1. Draw the graph of y = x 2 – 2x + 2 2. Find the axis of symmetry for the graph using 3. Then construct the table of values for x less than 1 and x greater than 1. 4. Graph the line y – 2x = -2 using slope-intercept form. 5. The points where the graphs intersect are the solution to the system. E XAMPLE 2 xy 0 1 2 3 (2,2) 5 2 1 2 5 y = x 2 – 2x + 2 y – 2x = -2 y = x 2 – 2x + 2 The quadratic-linear system has ______ solution(s). The solution(s) _________________ 1 (2, 2) +2 x + 2x y = 2x – 2 slope(m) = 2 y-int(b) = -2 y = x 2 – 2x + 2 Don’t forget to label the graphs. a = 1 b = -2 c = 2 I’m only using 5 points because my graph grid only goes up to 7 and -7. Show work for line in foldable, underneath box “Graph a line using…”

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Solve the quadratic-linear system graphically: 1. Draw the graph of y = x 2 – 2x + 1 2. Find the axis of symmetry for the graph using 3. Then construct the table of values for x less than 1 and x greater than 1. 4. Graph the line 3y = x - 6 using slope-intercept form. 5. The points where the graphs intersect are the solution to the system. E XAMPLE 3 xy 0 1 2 3 4 1 0 1 4 y = x 2 – 2x + 1 3y = x – 6 y = x 2 – 2x + 1 The quadratic-linear system has ______ solution(s). The solution(s) _________________ no none 3y = x – 6 3 3 3 y = 1x – 2 3 slope(m) = 1/3 y-int(b) = -2 y = x 2 – 2x + 1 Glue foldable on page 20 Don’t forget to label the graphs. a = 1 b = -2 c = 2 Show work for line in foldable, underneath box “Graph a line using…”

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Now in your notebook Page 21 Title it: Solving Quadratic-Linear Systems Algebraically 1.Solve: y = x 2 – 6x + 6 y – x = -4 (x 2 – 6x + 6) – x = -4 x 2 – 7x + 6 = -4 +4 +4 x 2 – 7x + 10 = 0 (x – 5 ) (x – 2) = 0 x = 5 x = 2 Steps: 1.Substitute “x 2 – 6x + 6” into the linear equation for “y”. 2.Solve for x. 3.Plug the value of x into either equation. ((I’m picking the linear equation) to get y. 4.Check with your calculator. y – x = -4 y – 5 = -4 + 5 +5 y = 1 y – x = -4 y – 2 = -4 + 2 +2 y = -2 Solution: (5, 1) and (2, -2) Now look back at #1 in the foldable that you just created. Compare the answers. Yay, they are the same.

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Page 22 2. Solve: y = x 2 – 2x + 2 y – 2x = -2 (x 2 – 2x + 2) – 2x = -2 x 2 – 4x + 2 = -2 +2 +2 x 2 – 4x + 4 = 0 Steps: 1.Substitute “x 2 – 2x + 2” into the linear equation for “y”. 2.Solve for x. 3.Plug the value of x into either equation. ((I’m picking the linear equation) to get y. 4.Check with your calculator. y – 2x = -2 y – 2(2) = -2 y – 4 = -2 + 4 +4 y = 2 Solution: (2, 2) a = 1 b = -4 c = 4 Can’t factor

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Page 23 3. Solve: y = x 2 – 2x + 1 3y = x – 6 3(x 2 – 2x + 1) = x – 6 3x 2 – 6x + 3 = x – 6 -x +6 -x +6 3x 2 – 7x + 9 = 0 Steps: 1.Substitute “x 2 – 2x + 1” into the linear equation for “y”. 2.Solve for x. Solution: no solution a = 3 b = -7 c = 9 Can’t factor Does not work therefore no solution

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What are three important characteristics of a parabola? Describe the two ways of finding the roots of a quadratic equation. What is one way to solve a system with a quadratic and a linear equation?

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