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.

EXAMPLE 1 Write a quadratic function in vertex form

A quadratic function for the parabola is y = (x – 1)2 – 2.

ANSWER

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.

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.

Solve for a.12

– = a

A quadratic function for the parabola is

12

– (x + 1)(x – 4) .y =

ANSWER

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.

EXAMPLE 3 Write a quadratic function in standard form

–3 = a(–1)2 + b(–1) + c Substitute –1 for x and 23 for y.

–3 = a – b + c Equation 1

–3 = 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

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.

STEP 2

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 3Solve the system consisting of revised Equations 1 and 3. Use the elimination method.

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

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)y = (x – 4)2 – 5

ANSWER2. vertex: (–3, 1)

passes through: (0, –8)y = (x + 3)2 + 1

ANSWER

3. x-intercepts: –2, 5 passes through: (6, 2)

y = (x + 2)(x – 5)14ANSWER

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)

y = 4x2 – 2x – 1

y = x2 + x + 3. –512

76

5. (–2, –1), (0, 3), (4, 1)

6. (–1, 0), (1, –2), (2, –15)

y = 4x2 x + 3

ANSWER

ANSWER

ANSWER