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
A. complete the square and get $\displaystyle y=\frac{-(2x-7)^2}{4}+\frac{33}{4}$...the axis of symmetry is $\displaystyle x=\frac{7}{2}$...and the vertex is at $\displaystyle \bigg(\frac{7}{2},\frac{33}{4}\bigg)$
B.The secon one factors into $\displaystyle y=3\bigg(x-\frac{4}{3}\bigg)^2+\frac{2}{3}$
and I am gettign tired of this so Ill come back and finish if someone else doesnt
I am back for
C. the equation is$\displaystyle y=\frac{-(x-3)^2}{3}+\frac{3}{2}$..the axis of symmetry is $\displaystyle x=3$...and the vertex is $\displaystyle \bigg(3,\frac{3}{2}\bigg)$
and for b part two the axis of symmetry is $\displaystyle x=\frac{4}{3}$...and the vertex is at $\displaystyle \bigg(\frac{4}{3},\frac{2}{3}\bigg)$
If I had $\displaystyle x^2-4x+13$ you could factor that if it was $\displaystyle x^2-4x+4=(x-2)^2$...but you dont have that 4...or do you...you can rewrite it it as $\displaystyle x^2-4x+4+9$ then factor the one part to get your final asnwer as $\displaystyle (x-2)^2+9$ 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 $\displaystyle x-2=0,x=2$ now put it in line form $\displaystyle x=2$ is the axis of symmetry....to find the vertex it is jsut $\displaystyle a(x-h)^2+k$ $\displaystyle (h,k)$ so once you get it in that order its simple