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.
Answer:
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)
