# Thread: completing the square help

1. ## completing the square help

Find the axis of symmetry and coordinates of vertex of the graph of the following quadratic functions using the method of completing the square.

A, y=-x^2+7x-4

B, y=3x^2-8x+6

C, y= -1/3x^2+2x-3/2

2. ## Ok

A. complete the square and get $y=\frac{-(2x-7)^2}{4}+\frac{33}{4}$...the axis of symmetry is $x=\frac{7}{2}$...and the vertex is at $\bigg(\frac{7}{2},\frac{33}{4}\bigg)$

B.The secon one factors into $y=3\bigg(x-\frac{4}{3}\bigg)^2+\frac{2}{3}$

and I am gettign tired of this so Ill come back and finish if someone else doesnt

3. ## Ok

I am back for

C. the equation is $y=\frac{-(x-3)^2}{3}+\frac{3}{2}$..the axis of symmetry is $x=3$...and the vertex is $\bigg(3,\frac{3}{2}\bigg)$

and for b part two the axis of symmetry is $x=\frac{4}{3}$...and the vertex is at $\bigg(\frac{4}{3},\frac{2}{3}\bigg)$

4. thanks for the answers but if it isn't too much to ask can you please explain the steps b/c i'm lost.

5. ## Here is what you do

If I had $x^2-4x+13$ you could factor that if it was $x^2-4x+4=(x-2)^2$...but you dont have that 4...or do you...you can rewrite it it as $x^2-4x+4+9$ then factor the one part to get your final asnwer as $(x-2)^2+9$ now that you have it in that form to find the axis of symmetry it is just the value that makes the inside of the parenthases zero...so in this case $x-2=0,x=2$ now put it in line form $x=2$ is the axis of symmetry....to find the vertex it is jsut $a(x-h)^2+k$ $(h,k)$ so once you get it in that order its simple