Find the quadratic equation whose parabola has vertex (3, –2) and y-intercept (0, 16). Give your answer in vertex form.

y = 2(x - 3)² - 2

Step-by-step explanation:

Given the quadratic equation with a y-intercept of (0, 16), and has a vertex occurring at point (3, -2):

We can substitute these values into the following vertex form of the quadratic equation:

y = a(x - h)² + k

where:

(h, k) = vertex

a = determines the wideness or narrowness of the graph, and the direction of where the parabola opens.

Now that we established the definitions for the variables in the vertex form, we can finally substitute the vertex, (3, -2) and y-intercept, (0, 16) into the vertex form and solve for the value of a:

y = a(x - h)² + k

16 = a(0 - 3)² - 2

16 = a(-3)² - 2

16 = 9a - 2

Add 2 to both sides:

16 + 2 = 9a - 2 + 2

18 = 9a

Divide both sides by 9 to solve for a:

2 = a

Therefore, the quadratic equation in vertex form is:

y = 2(x - 3)² - 2

Attached is a screenshot of the graphed equation, where it shows the y-intercept and the vertex.  

step-by-step explanation:

sry but any answer choices

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