Question

Write the equation of the parabola that has the vertex at point (3,−12) and passes through the point (0,6).

Answer 1

Answer:

y = 2(x - 3)² - 12

y = (-4/9)(x - 2)² + 7

Step-by-step explanation:

y = a(x - h)² + k

A) y = a(x - 3)² - 12

6 = a(0 - 3)² - 12

18 = 9a

a = 2

y = 2(x - 3)² - 12

B) y = a(x - 2)² + 7

3 = a(-1 - 2)² + 7

-4 = 9a

a = -4/9

y = (-4/9)(x - 2)² + 7

Answer 2

Answer:

y = 2(x-3)^2 -12

y = -4/9(x-2)^2 +7  bonus

Step-by-step explanation:

The vertex form of a parabola is

y = a(x-h)^2 + k  where (h,k) is the vertex

y = a(x-3)^2 - 12

We have one point given (0,6)

6 = a (0-3) ^2 -12

6 = a (-3)^2 -12

6 = 9a-12

Add 12 to each side

6+12 = 9a

18 = 9a

Divide each side by 9

18/9 = 9a/9

a=2

y = 2(x-3)^2 -12

We follow the same steps for the bonus

y = a(x-2)^2 +7

Substitute the point into the equation

3 = a (-1-2)^2 +7

3 =a (-3)^2 +7

3 = 9a +7

subtract 7 from each side

3-7 = 9a +7-7

-4 = 9a

Divide by 9

-4/9 =a

y = -4/9(x-2)^2 +7