1. 5.8 Modeling with Quadratic Functions

2. Modeling with Quadratic Functions
Depending on what information we are given will determine the form that is the most convenient to use.

3. Graphing y=x2 Standard Vertex y = ax2+bx+c y= a (x - h)2+k axis:
axis: x = h
vertex ( x , y )
vertex ( h , k )
C is y-intercept
a>0 U shaped, Minimum
a<0 ∩ shaped, Maximum

4. 1. Write a quadratic function for the parabola shown.
Use vertex form because the vertex is given.
y = a(x – h)2 + k
Vertex form
y = a(x – 1)2 – 2
Substitute 1 for h and –2 for k.
Use the other given point, (3, 2), to find a.
2 = a(3 – 1)2 – 2
Substitute 3 for x and 2 for y.
2 = 4a – 2
Simplify coefficient of a.
1 = a
Solve for a.
A quadratic function for the parabola is y = 1(x – 1)2 – 2.

5. Use vertex form because the vertex is given. y = a(x – h)2 + k
2. Write a quadratic function whose graph has the given characteristics.
vertex: (4, –5)
passes through: (2, –1)
Use vertex form because the vertex is given.
y = a(x – h)2 + k
Vertex form
y = a(x – 4)2 – 5
Substitute 4 for h and –5 for k.
Use the other given point, (2,–1), to find a.
–1 = a(2 – 4)2 – 5
Substitute 2 for x and –1 for y.
–1 = 4a – 5
Simplify coefficient of x.
1 = a
Solve for a.
A quadratic function for the parabola is y = 1(x – 4)2 – 5.
ANSWER

6. Use vertex form because the vertex is given.
passes through: (0, –8)
Use vertex form because the vertex is given.
y = a(x – h)2 + k
Vertex form
y = a(x + 3)2 + 1
Substitute –3 for h and 1 for k.
Use the other given point, (0,–8), to find a.
–8 = a(0 + 3)2 + 1
Substitute 2 for x and –8 for y.
–8 = 9a + 1
Simplify coefficient of x.
–1 = a
Solve for a.
A quadratic function for the parabola is y = 1(x + 3)2 + 1.
ANSWER

7. Steps for solving in 3 variables
Using the 1st 2 equations, cancel one of the variables.
Using the last 2 equations, cancel the same variable from step 1.
Use the results of steps 1 & 2 to solve for the 2 remaining variables.
Plug the results from step 3 into one of the original 3 equations and solve for the 3rd remaining variable.
Write the quadratic equation in standard form.

8. 4. Write a quadratic function in standard form for the parabola that passes through the points
STEP 1
Substitute the coordinates of each point into y = ax2 + bx + c to obtain the system of three linear equations shown below.
–3 = a(–1)2 + b(–1) + c
Substitute –1 for x and -3 for y.
–3 = a – b + c
Equation 1
–4 = a(0)2 + b(0) + c
Substitute 0 for x and – 4 for y.
– 4 = c
Equation 2
6 = a(2)2 + b(2) + c
Substitute 2 for x and 6 for y.
6 = 4a + 2b + c
Equation 3

9. STEP 2
Rewrite the system of three equations in Step 1 as a system of two equations by substituting – 4 for c in Equations 1 and 3.
a – b + c = – 3
Equation 1
a – b – 4 = – 3
Substitute – 4 for c.
a – b = 1
Revised Equation 1
4a + 2b + c = 6
Equation 3
4a + 2b - 4 = 6
Substitute – 4 for c.
4a + 2b = 10
Revised Equation 3

10. STEP 3
Solve the system consisting of revised Equations 1 and 3. Use the elimination method.
a – b = 1
2a – 2b = 2
4a + 2b = 10
4a + 2b = 10
6a = 12
a = 2
So 2 – b = 1, which means b = 1.
The solution is a = 2, b = 1, and c = – 4.
A quadratic function for the parabola is y = 2x2 + x – 4.
ANSWER

11. 5. Write a quadratic function in standard form for the parabola that passes through the given points.
(–1, 5), (0, –1), (2, 11)
STEP 1
Substitute the coordinates of each point into y = ax2 + bx + c to obtain the system of three linear equations shown below.
5 = a(–1)2 + b(–1) + c
Substitute –1 for x and 5 for y.
5 = a – b + c
Equation 1
–1 = a(0)2 + b(0) + c
Substitute 0 for x and – 1 for y.
– 1 = c
Equation 2
11 = a(2)2 + b(2) + c
Substitute 2 for x and 11 for y.
11 = 4a + 2b + c
Equation 3

12. STEP 2
Rewrite the system of three equations in Step 1 as a system of two equations by substituting – 1 for c in Equations 1 and 3.
a – b + c = 5
Equation 1
a – b – 1 = 5
Substitute – 1 for c.
a – b = 6
Revised Equation 1
4a + 2b + c = 11
Equation 3
4a + 2b – 1 = 11
Substitute – 1 for c.
4a + 2b = 12
Revised Equation 3

13. y = 4x2 – 2x – 1 STEP 3 Solve the system consisting of revised
Equations 1 and 3. Use the elimination method.
a – b = 6
2a – 2b = 12
4a + 2b = 10
4a + 2b = 12
6a = 24
a = 4
So, 4 – b = 6, which means b = – 2
A quadratic function for the parabola is
4x2 – 2x – 1
y =
ANSWER

14. 6. (1, 0), (2, -3), (3, -10)
STEP 1
Substitute the coordinates of each point into y = ax2 + bx + c to obtain the system of three linear equations shown below.
0 = a(1)2 + b(1) + c
Substitute 1 for x and 0 for y.
0= 1a + 1b + c
Equation 1
-3 = a(2)2 + b(2) + c
Substitute 2 for x and- 3 for y.
-3 = 4a +2b + c
Equation 2
-10 = a(3)2 + b(3) + c
Substitute 3 for x and -10 for y.
-10 = 9a + 3b + c
Equation 3

15. STEP 2 PART 1 -1 0 = 1a + 1b + c 0 = -1a - 1b - c -3 = 4a +2b + c
Rewrite the system of three equations in Step 1 as a system of two equations. Use Elimination to simplify one set of equations then use substitution & elimination to find the
PART 1 -1
0 = 1a + 1b + c
0 = -1a - 1b - c
Equation 1
-3 = 4a +2b + c
-3 = 4a +2b + c
Equation 2
-3 = 3a +1b
PART 1
PART 2 -1
3 = -4a -2b - c
-3 = 4a +2b + c
Equation 2
-10 = 9a + 3b + c
Equation 3
-10 = 9a + 3b + c
-7 = 5a + 1b
PART 2
-3 = 3a +1b -1
3 = -3a - 1b
PART 1
-7 = 5a + 1b
-7 = 5a + 1b
PART 2
-4 = 2a
-2 = a

16. STEP 3
Solve for the remaining variables by substitution.
Original equations
-3 = 3a +1b
0 = 1a + 1b + c
PART 1
-3 = 4a +2b + c
-7 = 5a + 1b
PART 2
-10 = 9a + 3b + c
-2 = a
-3 = 3a +1b
-3 = 3(-2) +1b
-3 = -6 +1b
3 = b
0 = 1a + 1b + c
0 = 1(-2) + 1(3) + c
0 = c
0 = 1 + c
-1 = c

17. y = -2x2 + 3x -1 Substitute y = ax2 + bx + c -2 = a 3 = b -1 = c
FINAL ANSWER!!