If a quadratic function has x-intercepts at 2 = - -3 and 2 = -
-1, one possible equation of the function would be y = (x + 3) (2+1), a graph that first
decreases until it reaches the vertex, (-3, -1), then increases (opening upward U shape).
So then, if a quadratic function has x-intercepts at 2 = -2 and 2 =
= 2
a. What is the equation of this quadratic function? Explain your reasoning
b. What is the vertex of the equation? Show your work and explain your reasoning.

Complete question is;

If a quadratic function has x-intercepts at x = -3 and x = -1, one possible equation of the function would be y = (x + 3)(x + 1), a graph that first

decreases until it reaches the vertex, (-3, -1), then increases (opening upward U shape).

So then, if a quadratic function has x-intercepts at x = -2 and x = 2

a. What is the equation of this quadratic function? Explain your reasoning

b. What is the vertex of the equation? Show your work and explain your reasoning.

Quadratic equation is; y = x² - 4

Vertex: (0, -4)

Step-by-step explanation:

The x-intercepts are at x = -2 and x = 2. Thus, the equation is;

y = (x - (-2))(x - 2)

y = x² + 2x - 2x - 4

y = x² - 4

Now,the coordinates of the vertex will be (h, k)

Where;

h = -b/2a

k is the value of y at x = h

From y = x² - 4, h = 0

Thus, putting x = h = 0 in the quadratic equation, we have;

y = 0² - 4

y = -4

Thus,the coordinates of the vertex are; (0, -4)

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step-by-step explanation:

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