# Notes: Graphing Quadratic Functions and solving quadratic linear systems algebraically Aim: Students will be able to solve quadratic linear systems graphically.

## Presentation on theme: "Notes: Graphing Quadratic Functions and solving quadratic linear systems algebraically Aim: Students will be able to solve quadratic linear systems graphically."— Presentation transcript:

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”.

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 linear quadratic linear solution set of ordered pairs intersects two one no 2 solutions 1 solution no solution

Example 1 Solve the quadratic-linear system graphically: y = x2 – 6x + 6 y – x = -4 1. Draw the graph of y = x2 – 6x 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 Graph the line y – x = -4 using slope-intercept form. 5. The points where the graphs intersect are the solution to the system. y = x2 – 6x + 6 Show work for line in foldable, underneath box “Graph a line using…” + x + x y = x – 4 x y 1 2 3 4 5 6 6 slope(m) = 1 y-int(b) = -4 a = 1 b = -6 c = 6 1 -2 -3 y = x2 – 6x + 6 -2 1 6 [(2,-2) and (5,1)] The quadratic-linear system has ______ solution(s). The solution(s) _________________ 2 Don’t forget to label the graphs. (2, -2) and (5, 1)

Check with graphing calculator
Type the 2 functions into your graphing calculator. y1 = x2 + 6x + 6 AND the line y2 = x – 4 on the interval [0,6] quadratic line x y1 y2 6 -4 1 -3 2 -2 3 -1 4 5 Solution to the system: {(-2, -2), (1, 1)}

Example 2 Solve the quadratic-linear system graphically: y = x2 – 2x + 2 y – 2x = -2 1. Draw the graph of y = x2 – 2x 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 Graph the line y – 2x = -2 using slope-intercept form. 5. The points where the graphs intersect are the solution to the system. Show work for line in foldable, underneath box “Graph a line using…” +2 x x y = 2x – 2 y = x2 – 2x + 2 x y -1 1 2 3 slope(m) = 2 y-int(b) = -2 a = 1 b = -2 c = 2 5 I’m only using 5 points because my graph grid only goes up to 7 and -7. 2 1 2 y = x2 – 2x + 2 5 (2,2) The quadratic-linear system has ______ solution(s). The solution(s) _________________ 1 Don’t forget to label the graphs. (2, 2)

Example 3 Solve the quadratic-linear system graphically: y = x2 – 2x + 1 3y = x – 6 Show work for line in foldable, underneath box “Graph a line using…” 1. Draw the graph of y = x2 – 2x 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 Graph the line 3y = x - 6 using slope-intercept form. 5. The points where the graphs intersect are the solution to the system. y = x2 – 2x + 1 x y -1 1 2 3 4 a = 1 b = -2 c = 2 1 y = x2 – 2x + 1 1 4 3y = x – 6 y = 1x – 2 3 Don’t forget to label the graphs. slope(m) = 1/3 y-int(b) = -2 The quadratic-linear system has ______ solution(s). The solution(s) _________________ no Glue foldable on page 20 none

Now in your notebook Page 21 Title it: Solving Quadratic-Linear Systems Algebraically Solve: y = x2 – 6x + 6 y – x = -4 Steps: Substitute “x2 – 6x + 6” into the linear equation for “y”. Solve for x. Plug the value of x into either equation. ((I’m picking the linear equation) to get y. Check with your calculator. (x2 – 6x + 6) – x = -4 x2 – 7x + 6 = -4 x2 – 7x + 10 = 0 (x – 5 ) (x – 2) = 0 x = x = 2 Now look back at #1 in the foldable that you just created. Compare the answers. y – x = -4 y – 5 = -4 y = 1 y – x = -4 y – 2 = -4 y = -2 Yay, they are the same. Solution: (5, 1) and (2, -2)

2. Solve: y = x2 – 2x + 2 y – 2x = -2 Page 22 Steps:
Substitute “x2 – 2x + 2” into the linear equation for “y”. Solve for x. Plug the value of x into either equation. ((I’m picking the linear equation) to get y. Check with your calculator. (x2 – 2x + 2) – 2x = -2 x2 – 4x + 2 = -2 x2 – 4x + 4 = 0 Can’t factor a = 1 b = -4 c = 4 y – 2x = -2 y – 2(2) = -2 y – 4 = -2 y = 2 Solution: (2, 2)

3. Solve: y = x2 – 2x + 1 3y = x – 6 Page 23 3(x2 – 2x + 1) = x – 6
Steps: Substitute “x2 – 2x + 1” into the linear equation for “y”. Solve for x. Can’t factor a = 3 b = -7 c = 9 Does not work therefore no solution Solution: no solution

3-2-1 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?