Question

calculating a vertex
consider the quadratic function:
$f(x)=x^2 - 8x - 9$
vertex: $left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\
ight)\
ight)$
what is the vertex of the function?
$(\quad, \quad)$

Explanation:

Step1: Identify a, b in quadratic function

For \( f(x) = x^2 - 8x - 9 \), the standard form of a quadratic function is \( f(x) = ax^2 + bx + c \). Here, \( a = 1 \), \( b = -8 \), \( c = -9 \).

Step2: Calculate x-coordinate of vertex

Using the formula \( x = \frac{-b}{2a} \), substitute \( a = 1 \), \( b = -8 \):
\( x = \frac{-(-8)}{2(1)} = \frac{8}{2} = 4 \).

Step3: Calculate y-coordinate of vertex

Substitute \( x = 4 \) into \( f(x) = x^2 - 8x - 9 \):
\( f(4) = (4)^2 - 8(4) - 9 = 16 - 32 - 9 = -25 \).

Answer:

\( (4, -25) \)