Question

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

Answer 1

Answer:

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.

• a > 1: the graph is narrower than the parent function; the parabola also opens upward.
• 0 < a < 1: the graph is wider than the parent function.
• a < 1: the graph opens downward.

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:

$\frac{18}{9} = \frac{9a}{9}$

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.