which function has a vertex at $(2, -9)$?\n\n$f(x) = -(x - 3)^2$\n$f(x) = (x + 8)^2$\n$f(x) = (x - 5)(x +…

Explanation:

Step 1: Identify vertex form and intercept form

The vertex form is $f(x) = a(x - h)^2 + k$ and intercept form is $f(x) = a(x - p)(x - q)$.

Step 2: Evaluate the first two options

For $f(x) = -(x - 3)^2$, vertex is $(3, 0)$. For $f(x) = (x + 8)^2$, vertex is $(-8, 0)$.

Step 3: Calculate vertex for the third option

For $f(x) = (x - 5)(x + 1)$, the $x$-coordinate is $h = \frac{5 + (-1)}{2} = 2$. $$f(2) = (2 - 5)(2 + 1) = (-3)(3) = -9$$

Step 4: Calculate vertex for the fourth option

For $f(x) = -(x - 1)(x - 5)$, the $x$-coordinate is $h = \frac{1 + 5}{2} = 3$. $$f(3) = -(3 - 1)(3 - 5) = -(2)(-2) = 4$$

Answer:

$f(x) = (x - 5)(x + 1)$