Answer:
The vertex is at [tex](0, -9)[/tex]
The axis of symmetry [tex]x = 0[/tex]
Step-by-step explanation:
Note: You can answer all the questions looking at the graph
We have the function [tex]f: \mathbb{R} \rightarrow\mathbb{R}[/tex](I am assuming this) such that [tex]f(x) = y = (x-3)(x+3)[/tex]
We note that
[tex](x-3)(x+3) = x^2-9[/tex] once we have a difference of squares
Therefore, we have a quadratic function.
So
[tex]y = x^2-9[/tex]
The vertex [tex](h, k)[/tex] is
[tex]h = \dfrac{-b}{2a} = \dfrac{-0}{2 \cdot 1} = 0[/tex]
[tex]k= h^2 - 9 = 0^2 - 9 =-9[/tex]
Therefore, the vertex is at [tex](0, -9)[/tex]
As we know the vertex, we conclude that the axis of symmetry [tex]x = 0[/tex]
The x-intercepts occur at [tex]y = 0[/tex]
[tex]y = (x-3)(x+3) \implies x = -3, x=3[/tex]
So, the left x-intercept is [tex](-3, 0)[/tex] and the right x-intercept is [tex](3, 0)[/tex]
As I stated in the beginning, the domain is the set of all real numbers. Precisely,
[tex]Dom(f) = \mathbb{R} = (-\infty, \infty)[/tex]
Once, we know the vertex, we also conclude that the Range is
[tex]Ran(f)= [-9,\infty)[/tex]
Greater than or equal to -9
