Question

Determine the vertex and the axis of symmetry for the function below: y =-3/4x^2+6x+6

Answer 1

Hmm so our leading term (largest power of x) is x^2, so we have a parabola.
The sign on the leading term is negative, telling us that it will open downward.
We'll have to get it into standard form so we can identify the vertex easier.

\[\Large y=-\frac{3}{4}x^2+6x+6\]

Answer 2

Lemme know if this step is too confusing.
We want a coefficient of 1 on our x^2.
So we'll factor -3/4 out of each term, giving us,\[\Large y=-\frac{3}{4}\left(x^2-8x-8\right)\]

Answer 3

Oh actually I guess we'll need to complete the square on the x's.. so it makes more sense to only factor -3/4's from the x's.\[\Large y=-\frac{3}{4}\left(x^2-8x\right)+6\]

Answer 4

Are the math equations showing up ok Kundi? c:
Confused by that last step? :o

Answer 5

a little

Answer 6

im not really sure how you got the 8 either

Answer 7

Factoring is a little tricky when we deal with fractions.\[\Large y=\left(-\frac{3}{4}x^2+6x\right)+6\]The reason I put brackets here is just so you know that we're only dealing with these two terms right now.Factoring a -3/4 out of each term means we'll multiply the outside of the brackets by -3/4 while we divide each term by -3/4 on the inside.

\[\Large y=-\frac{3}{4}\left(\frac{-\dfrac{3}{4}x^2}{-\frac{3}{4}}+\frac{6x}{-\frac{3}{4}}\right)+6\]

Answer 8

That might make it more confusing lol I dunno..
You can let me know :p

Answer 9

When we divide by a fraction, we can rewrite it as multiplication by `flipping` the bottom fraction. And we also have the -3/4 cancel out in the first fraction.
That was the point of factoring the value out.\[\Large y=-\frac{3}{4}\left(\frac{\cancel{-\dfrac{3}{4}}x^2}{\cancel{-\frac{3}{4}}}+6x\left(-\frac{4}{3}\right)\right)+6\]

Answer 10

\[\Large y=-\frac{3}{4}\left(x^2-8x\right)+6\]

Answer 11

oooo ok thanks.