Question

complete the square on the quadratic state the vertex, min or max values, axis of symmetry
y=-2x^2+3x+6

Answer 1

\[y=-2(x-\frac{3}{4})^2+\text { something}\] and we can find the something easy enough

Answer 2

since
\[(x-\frac{3}{4})^2=0\iff x=\frac{3}{4}\] replace x by 3/4 in your original expression to find the something.
then because this si a parabola that opens down, it will be a maximum.

Answer 3

i believe it is
\[\frac{57}{8}\] but you should check my arithmetic

Answer 4

this is the max correct? how do I find the square.

Answer 5

you find the vertex via
\[x=-\frac{b}{2a}\] in this case you have
\[a=-2,b=3,-\frac{b}{2a}=\frac{3}{4}\] so you know it will look like
\[y=-2(x-\frac{3}{4})^2+k\] and you find k by substitution

Answer 6

ok

Answer 7

there is a harder way to do this, but if you can remember
\[-\frac{b}{2a}\] (which is not so bad) then you will be in good shape

Answer 8

ok thanks