1. Write a quadratic function in vertex form
EXAMPLE 1
Write a quadratic function in vertex form
Write a quadratic function for the parabola shown.
SOLUTION
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.

2. EXAMPLE 1
Write a quadratic function in vertex form
ANSWER
A quadratic function for the parabola is y = (x – 1)2 – 2.

3. Write a quadratic function in intercept form
EXAMPLE 2
Write a quadratic function in intercept form
Write a quadratic function for the parabola shown.
SOLUTION
Use intercept form because the x-intercepts are given.
y = a(x – p)(x – q)
Intercept form
y = a(x + 1)(x – 4)
Substitute –1 for p and 4 for q.

4. Write a quadratic function in intercept form
EXAMPLE 2
Write a quadratic function in intercept form
Use the other given point, (3, 2), to find a.
2 = a(3 + 1)(3 – 4)
Substitute 3 for x and 2 for y.
2 = – 4a
Simplify coefficient of a. 1 2 – = a
Solve for a.
A quadratic function for the parabola is 1 2 –
(x + 1)(x – 4) .
y =
ANSWER

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

6. Write a quadratic function in standard form
EXAMPLE 3
Write a quadratic function in standard form
–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
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.

7. Write a quadratic function in standard form
EXAMPLE 3
Write a quadratic function in standard form
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
STEP 3
Solve the system consisting of revised Equations 1 and 3. Use the elimination method.

8. EXAMPLE 3
Write a quadratic function in standard form
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

9. GUIDED PRACTICE
for Examples 1, 2 and 3
Write a quadratic function whose graph has the given characteristics.
1. vertex: (4, –5)
passes through: (2, –1)

10. Use vertex form because the vertex is given. y = a(x – h)2 + k
GUIDED PRACTICE
for Examples 1, 2 and 3
SOLUTION
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 = (x – 4)2 – 5.
ANSWER

11. GUIDED PRACTICE
for Examples 1, 2 and 3
2. vertex: (–3, 1)
passes through: (0, –8)

12. Use vertex form because the vertex is given. y = a(x – h)2 + k
GUIDED PRACTICE
for Examples 1, 2 and 3
SOLUTION
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,–1), to find a.
–8 = a(0 + 3)2 + 1
Substitute 0 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 = (x + 3)2 + 1.
ANSWER

13. Use intercept form because the x-intercepts are given.
GUIDED PRACTICE
for Examples 1, 2 and 3
3. x-intercepts: –2, 5
passes through: (6, 2)
SOLUTION
Use intercept form because the x-intercepts are given.
y = a(x – p)(x – q)
Intercept form
y = a(x - (– 2))(x – 5)
Substitute –2 for p and –5 for q.

14. Use the other given point, (6, 2), to find a.
GUIDED PRACTICE
for Examples 1, 2 and 3
Use the other given point, (6, 2), to find a.
2 = a(6 + 2)(6 - 5)
Substitute 6 for a and 2 for y.
2 = 8a
Simplify coefficient of a. 1 4 = a
Solve for a.
A quadratic function for the parabola is 1 4
(x + 2)(x – 5) .
y =
ANSWER

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

16. GUIDED PRACTICE for Examples 1, 2 and 3 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
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.

17. GUIDED PRACTICE for Examples 1, 2 and 3 a – b + c = 5 a – b – 1 = 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
STEP 3
Solve the system consisting of revised Equations 1 and 3. Use the elimination method.

18. GUIDED PRACTICE
for Examples 1, 2 and 3
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

19. GUIDED PRACTICE
for Examples 1, 2 and 3
5. (–2, –1), (0, 3), (4, 1)
SOLUTION
STEP 1
Substitute the coordinates of each point into y = ax2 + bx + c = 0 to obtain the system of three linear equations shown below.

20. GUIDED PRACTICE for Examples 1, 2 and 3 –1 = a(–2)2 + b(–2) + c
Substitute –2 for y and –1 for x.
–1 = 4a – 2b + c
Equation 1
3 = a(0)2 + b(0) + c
Substitute 0 for x and 3 for y.
3 = c
Equation 2
1 = a(4)2 + b(4) + c
Substitute 1 for y and 4 for x.
1 = 16a + 4b + c
Equation 3
STEP 2
Rewrite the system of three equations in Step 1 as a system of two equations by substituting 3 for c in Equations 1 and 3.

21. GUIDED PRACTICE for Examples 1, 2 and 3 –1 = 4a – 2b + c
Equation 1
4a – 2b + 3 = – 1
Substitute 3 for c.
4a – 2b = – 1
Revised Equation 1
1 = 16a + 4b + c
Equation 3
16a + 4b + 3 = 1
Substitute 3 for c.
16a2 + 4b = –2
Revised Equation 3
STEP 3
Solve the system consisting of revised Equations 1 and 3. Use the elimination method.

22. GUIDED PRACTICE
for Examples 1, 2 and 3
4a – 2b = – 4
8a – 4b = – 8
16a + 4b = – 2
16a + 4b = – 2
24a = –10
a = 5 12
So b = , which means b = 5 12 7 6
ANSWER
A quadratic function for the parabola is y = x x + 3.
– 5 12 7 6

23. GUIDED PRACTICE
for Examples 1, 2 and 3
6. (–1, 0), (1, –2), (2, –15)
SOLUTION
STEP 1
Substitute the coordinates of each point into y = ax2 + bx + c to obtain the system of three linear equations shown below.

24. GUIDED PRACTICE
for Examples 1, 2 and 3
ANSWER
A quadratic function for the parabola is y = 4x2  x + 3