SOLUTION: convert quadratic polynomial y(x)=x^2-6x+7 into vertex form [completeting full sqaure];find Xv,Yv, y-intercept, x-intercept and graph it.

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Question 117705: convert quadratic polynomial y(x)=x^2-6x+7 into vertex form [completeting full sqaure];find Xv,Yv, y-intercept, x-intercept and graph it.
Found 2 solutions by josmiceli, edjones:
Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!

Find roots

complete the square by taking half of the , square it,
then add it to both sides

The roots (x-intercepts) are at (3 + , 0) and (3 - ,0)
The vertex is exactly between them at (3, y). What is ?
make

So, the vertex is at (3,-2)
Lastly, where is the y-intercept? It is where

The y-intercept is at (0,7)
Here's the graph

Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website!
y(x)=x^2-6x+7
=x^2-6x+9-9+7 complete the square.
=(x-3)^2-2
y=a(x-h)^2+k
a=1, h=3, k=-2
Vertex=(h,k)=(3,-2)=(Xv,Yv)
.
the y intercept=y(0)=7
.
the x intercepts=
x^2-6x+7=0
(x-3)^2-2=0 See above.
(x-3)^2=2
x-3=+-sqrt(2) Take sqrt of both sides.
x=3+-sqrt(2) the x intercepts.
.
Ed