To find which function has a vertex at (2, -9), we can analyze each option and determine the vertex of each quadratic function.
-
f(x) = –(x – 3)²: This function is in vertex form \( f(x) = a(x-h)² + k \), where \( (h, k) \) is the vertex. Here, \( h = 3 \) and \( k = 0 \). So, the vertex is (3, 0).
-
f(x) = (x + 8)²: In vertex form, here \( h = -8 \) and \( k = 0 \), giving the vertex (–8, 0).
-
f(x) = (x – 5)(x + 1): To find the vertex, we first need to expand this: \[ f(x) = x² + x - 5x - 5 = x² - 4x - 5. \] To find the vertex of this quadratic, we can use the formula \( x = -\frac{b}{2a} \). Here \( a = 1 \) and \( b = -4 \): \[ x = -\frac{-4}{2(1)} = 2. \] Now we plug \( x = 2 \) back into the function to find \( y \): \[ f(2) = (2 - 5)(2 + 1) = (-3)(3) = -9. \] So, the vertex is (2, -9).
-
f(x) = –(x – 1)(x – 5): Expand this function: \[ f(x) = -[x² - 6x + 5] = -x² + 6x - 5. \] The vertex can be found using \( x = -\frac{b}{2a} \) with \( a = -1 \) and \( b = 6 \): \[ x = -\frac{6}{2(-1)} = 3. \] Now substitute \( x = 3 \) back to find \( y \): \[ f(3) = -3² + 6(3) - 5 = -9 + 18 - 5 = 4. \] So, the vertex is (3, 4).
The only function that has a vertex at (2, -9) is:
\[ \boxed{f(x) = (x - 5)(x + 1)}. \]