Question

What is the vertex of the function?

Answer 1

Answer:

  (4, -25)

Step-by-step explanation:

One way to answer this is to put the equation into vertex form.

  f(x)= x^2 -8x -9 . . . . . given

Add and subtract the square of half the x-coefficient:

  f(x) = x^2 -8x +(-8/2)^2 -9 -(-8/2)^2

  f(x) = x^2 -8x +16 -25

  f(x) = (x -4)^2 -25

Comparing this to the vertex form of a quadratic:

  f(x) = a(x -h)^2 +k

we find that (h, k) = (4, -25). This is the vertex.

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Alternate solution

The line of symmetry for ...

  f(x) = ax^2 +bx +c

is given by x = -b/(2a). For your given quadratic that line is ...

  x = -(-8)/(2(1)) = 4

Evaluating f(4) gives the y-coordinate:

  f(4) = 4^2 -8·4 -9 = -25

The vertex is (x, y) = (4, -25).