anonymous
  • anonymous
Determine the vertex and the axis of symmetry for the function below: y =-3/4x^2+6x+6
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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zepdrix
  • zepdrix
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. \[\Large y=-\frac{3}{4}x^2+6x+6\]
zepdrix
  • zepdrix
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,\[\Large y=-\frac{3}{4}\left(x^2-8x-8\right)\]
zepdrix
  • zepdrix
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.\[\Large y=-\frac{3}{4}\left(x^2-8x\right)+6\]

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zepdrix
  • zepdrix
Are the math equations showing up ok Kundi? c: Confused by that last step? :o
anonymous
  • anonymous
a little
anonymous
  • anonymous
im not really sure how you got the 8 either
zepdrix
  • zepdrix
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. \[\Large y=-\frac{3}{4}\left(\frac{-\dfrac{3}{4}x^2}{-\frac{3}{4}}+\frac{6x}{-\frac{3}{4}}\right)+6\]
zepdrix
  • zepdrix
That might make it more confusing lol I dunno.. You can let me know :p
zepdrix
  • zepdrix
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.\[\Large y=-\frac{3}{4}\left(\frac{\cancel{-\dfrac{3}{4}}x^2}{\cancel{-\frac{3}{4}}}+6x\left(-\frac{4}{3}\right)\right)+6\]
zepdrix
  • zepdrix
\[\Large y=-\frac{3}{4}\left(x^2-8x\right)+6\]
anonymous
  • anonymous
oooo ok thanks.

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