Question is, find the vertex of the parabola...

f(x)= x

^{2}-8x+23-
CREATE FREE ACCOUNT
- Access thousands of free resources from expert tutors
- Comment on posts and interact with the authors
- Ask questions and get free answers from tutors
- View videos, take interactive quizzes, and more!

- Become a Student
- Become a Student
- Sign In

Question is, find the vertex of the parabola...

f(x)= x^{2}-8x+23

Tutors, please sign in to answer this question.

Hi, Carlos.

There are two ways to approach this problem. We need to put this equation into a different form.

f(x) = a(x-h)^{2} + k, so the vertex will be the point (h, k).

One way is to complete the square: f(x) = (x^{2} - 8x + ) + 23.

What will make the perfect square? Take half of 8, square it , and add it to the inside and subtract from the outside on the parenthesis.

This results in f(x) = (x^{2} - 8x + 16) + 23 - 16.

And f(x) = (x - 4)^{2} + 7. Therefore, the axis of symmetry is x = 4, and the vertex is the point (4, 7)

Alternatively, you can find the value of f, by using the equation x = -b/(2a), where b is the coefficient of x and a is the coefficient on x^{2}. This results in x = -(-8)/(2*1).

So the x value of the vertex is 4 and the y value can be found by finding f(4).

f(4) = 4^{2} - 8*4 + 23 = 16 - 32 + 23 = 7. So the vertex is (4, 7).

I hope this helps.

Jamie S.

Licensed Spanish Teacher - All Levels!

Gainesville, VA

5.0
(148 ratings)

Pnina M.

Patient, Experienced Tutor, Great Results!

Fairfax, VA

4.8
(4 ratings)

ROLF S.

Older German available for tutoring in greater Sterling area

Sterling, VA

3.0
(1 ratings)