##### Find the vertex and give the equation for axis symmetry.

 Algebra Tutor: None Selected Time limit: 1 Day

f(x) = (x-7)^2+5

Sep 9th, 2015

Wanna,

I'd suggest starting with a plot, just a sketch, really. If you do it, you see that the graph of f(x) is a parabola that opens up and that when x = 7, f(x) = 5. The vertex, then is the point (7,5). You can see that geometrically. f(x) is symmetric about the (vertical) line x = 7. You can see it algebraically by noting that, for example,

f(6) = (6-7)^2 + 5 = 6 = (8-7)^2 + 5. More generally, f(x) = f(-x).

Sep 9th, 2015

I am still at a lost at how you got the answer. Is there some steps that might assist me?

Sep 9th, 2015

Can you tell me at what point my explanation became unclear? I can also ask you a few questions to help narrow things down. Your choice.

Sep 9th, 2015

Well in class we were given a formula to find vertex so, I assume it is not needed in this question? Is it possible to to find the vertex without graphing first?

Sep 9th, 2015

Please type in the formula you were given. We'll go from there.

Sep 9th, 2015

-b/2a

Sep 9th, 2015

Ok. Thx. Can U tell me what the 'a' and 'b' are related to? It should be an equation.

Sep 9th, 2015

The vertex of the parabola  y= ax^2  +bx + c has x-coordinate -b/2(a). Thy y-coordinate 4(a)(c)-b^2/4(a)

Sep 9th, 2015

Ok. The equation y= f(x) = ax^2  +bx + c is what I was hoping you'd bring. Can U make the equation
f(x) = (x-7)^2+5 look like ax^2 +bx + c?

Sep 9th, 2015

No, well at least I wouldnt know how if you could.

Sep 9th, 2015

Try carrying out the indicated operations: (x-7)^2 + 5. What do you get?

Sep 9th, 2015

x^2+49+5....or should it be (x-7)(x-7)+5    x^2-7x-7x+49 = x^2+49+5......which is the same, unless im doing something wrong.

Sep 9th, 2015

You correctly got (x-7)^2 = x^2-7x-7x+49 = x^2-14x + 49. So, f(x) = x^2 - 14x + 54. In this case, we have

-b/2a = -(-14)/(2*1) = 14/2 = 7. The equation of the axis of symmetry is (as I noted) x = -b/2a = 7.

I need to leave now. I need to attend to other students. Best wishes. Let me know if I can help in the future (on a different problem). Bye.

Sep 9th, 2015

Okay thank you

Sep 9th, 2015

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Sep 9th, 2015
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Sep 9th, 2015
May 24th, 2017
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