Question

y=x²-10x+7

Answer 1

Nice, what is the question?

Answer 2

write the quadratic function vertex form and identify the vertex

Answer 3

So to get it in vertex form you want to get it to look like this:
\(y=(x-a)^2+b\)
Since that describes a parabola shifted \(a\) units to the right and \(b\) units upwards, so that the vertex is at \((a,b)\). To do that you need to complete the square.

Answer 4

yes and do i half the 10x

Answer 5

Exactly, you put half of the coefficient of \(x\) as \(a\). Since \((x-a)^2=x^2-2ax+a^2\) you see that that \(-2a\) will be the same as the coefficient in front of \(x\).

Answer 6

so half of ten is 5

Answer 7

Yes, so the completed square gives \((x-5)^2=x^2-10x+5^2\), that almost looks like what you started with, but with \(5^2=25\) instead of \(7\). So the trick is to still write down the completed square \((x-5)^2\), but subtract from it so that you don't get 25, but 7.

Answer 8

SO IT WILL LOOK LIKE THIS (X²-5)²

Answer 9

Yes, but without the 2 over the x. You also need to subtract 18 to get 7, so it matches what you started with. In other words,
\[x^2-10x+7=(x-5)^2-18\]

Answer 10

Now compare that with the general equation I posted above (with the a's and b's) to find the vertex.