EXAMPLE 1

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# EXAMPLE 1 - PowerPoint PPT Presentation

Write a quadratic function in vertex form. EXAMPLE 1. 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.

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Write a quadratic function in vertex form

EXAMPLE 1

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 forhand –2 for k.

Use the other given point, (3, 2), to find a.

2= a(3– 1)2 – 2

Substitute 3 for xand 2 for y.

2 = 4a – 2

Simplify coefficient of a.

1 = a

Solve for a.

Write a quadratic function in vertex form

EXAMPLE 1

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

Write a quadratic function in intercept form

EXAMPLE 2

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 pand 4 for q.

1

1

a

=

2

2

A quadratic function for the parabola is

y =

(x + 1)(x – 4) .

Write a quadratic function in intercept form

EXAMPLE 2

Use the other given point, (3, 2), to find a.

2= a(3+ 1)(3– 4)

Substitute 3 for xand 2 for y.

2 = –4a

Simplify coefficient of a.

Solve for a.

Write a quadratic function in standard form

EXAMPLE 3

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 + cto obtain the system of three linear equations shown below.

Write a quadratic function in standard form

EXAMPLE 3

–3= a(–1)2 + b(–1) + c

Substitute –1 for xand 23 for y.

–3 = a – b + c

Equation 1

–3= a(0)2 + b(0) + c

Substitute 0 for xand –4 for y.

–4 = c

Equation 2

6= a(2)2 + b(2) + c

Substitute 2 for xand 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 cin Equations 1 and 3.

Write a quadratic function in standard form

EXAMPLE 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

STEP 3

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

6a = 12

A quadratic function for the parabola is y = 2x2 + x – 4.

Write a quadratic function in standard form

EXAMPLE 3

2a – 2b = 2

a – b = 1

4a + 2b = 10

4a + 2b = 10

a = 2

So 2 – b = 1, which means b = 1.

The solution is a = 2, b = 1, and c = –4.

1. vertex: (4, –5)

1

passes through:(2, –1)

4

2. vertex: (–3, 1)

passes through: (0, –8)

3.x-intercepts: –2, 5

passes through: (6, 2)

y = (x + 2)(x – 5)

for Examples 1, 2 and 3

GUIDED PRACTICE

Write a quadratic function whose graph has the given characteristics.

y = (x – 4)2 – 5

y = (x + 3)2 + 1

7

6

–5

12

for Examples 1, 2 and 3

GUIDED PRACTICE

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

5. (–2, –1), (0, 3), (4, 1)

y = x2 + x + 3.

6. (–1, 0), (1, –2), (2, –15)

y = 4x2x + 3