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

f(x)= x

^{2}-8x+23-
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Question is, find the vertex of the parabola...

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

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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.

Susana M.

Spanish, ESL and English Tutoring

Colonia, NJ

Jonathan F.

Current AP Calculus Teacher and University Professor

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Kathrin P.

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