1. 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”.

2. What is a quadratic-linear 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 linear
quadratic
linear
solution
set of ordered pairs
intersects
two
one no
2 solutions
1 solution
no solution

3. Solve the quadratic-linear system graphically:
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)

4. 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)}

5. Solve the quadratic-linear system graphically:
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)

6. Solve the quadratic-linear system graphically:
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

7. 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)

8. 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)

9. 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

10. 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?