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Where is the vertex of the parabola? y = x2 + 2x – 3 a. above the x-axis b. below the x-axis c. on the x-axis d. on the y-axis
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complete the square and decide. f(x)=a(x-h)^2+k if k=0, vertex is on the x-axis, k>0 it is above if h=0, vertex is on the y-axis, h<0 it is to the right.
I tried to do that, I can't figure it out
I can give an example: For \(x^2+4x-3\) \(x^2+4x-3\) =\(x^2+2(2x)+4-4-3\) =\((x^2+2(2x)+4)-7\) notice the perfect square on the left! =\((x+2)^2-7\) compare with f(x)=a(x-h)^2+k so a=1, h=-2, k=-7 so the vertex (of this example) is to the left of the y-axis, and below the x-axis.
thank you! I understand a lot better now!
You're welcome! :)
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