Answer :
To find the vertex of the quadratic function \( f(x) = x^2 - 8x - 9 \), we will use the vertex formula for a parabola given by \( f(x) = ax^2 + bx + c \):
The formula for the x-coordinate of the vertex is:
[tex]\[ x = \frac{-b}{2a} \][/tex]
In our function, we identify \( a = 1 \), \( b = -8 \), and \( c = -9 \).
First, we calculate the x-coordinate of the vertex:
[tex]\[ x = \frac{-(-8)}{2 \cdot 1} = \frac{8}{2} = 4 \][/tex]
Next, we need to find the y-coordinate of the vertex by substituting \( x = 4 \) back into the original function \( f(x) \):
[tex]\[ f(4) = (4)^2 - 8(4) - 9 \][/tex]
We calculate each term step-by-step:
[tex]\[ (4)^2 = 16 \][/tex]
[tex]\[ -8(4) = -32 \][/tex]
So plugging these back into the function:
[tex]\[ f(4) = 16 - 32 - 9 \][/tex]
Now, we simplify this:
[tex]\[ 16 - 32 = -16 \][/tex]
[tex]\[ -16 - 9 = -25 \][/tex]
Therefore, the y-coordinate of the vertex is:
[tex]\[ f(4) = -25 \][/tex]
Hence, the vertex of the function \( f(x) = x^2 - 8x - 9 \) is:
[tex]\[ (4, -25) \][/tex]
So, the vertex of the function is [tex]\(\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right) = (4, -25)\)[/tex].
The formula for the x-coordinate of the vertex is:
[tex]\[ x = \frac{-b}{2a} \][/tex]
In our function, we identify \( a = 1 \), \( b = -8 \), and \( c = -9 \).
First, we calculate the x-coordinate of the vertex:
[tex]\[ x = \frac{-(-8)}{2 \cdot 1} = \frac{8}{2} = 4 \][/tex]
Next, we need to find the y-coordinate of the vertex by substituting \( x = 4 \) back into the original function \( f(x) \):
[tex]\[ f(4) = (4)^2 - 8(4) - 9 \][/tex]
We calculate each term step-by-step:
[tex]\[ (4)^2 = 16 \][/tex]
[tex]\[ -8(4) = -32 \][/tex]
So plugging these back into the function:
[tex]\[ f(4) = 16 - 32 - 9 \][/tex]
Now, we simplify this:
[tex]\[ 16 - 32 = -16 \][/tex]
[tex]\[ -16 - 9 = -25 \][/tex]
Therefore, the y-coordinate of the vertex is:
[tex]\[ f(4) = -25 \][/tex]
Hence, the vertex of the function \( f(x) = x^2 - 8x - 9 \) is:
[tex]\[ (4, -25) \][/tex]
So, the vertex of the function is [tex]\(\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right) = (4, -25)\)[/tex].
