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for what values of a and b will the vertex of the graph of y=ax2 +bx be located at (-1, -2)?
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The vertex is (h,k) = (-1,-2), so h = -1 and k = -2 y = a(x-h)^2 + k y = a(x-(-1))^2 - 2 y = a(x+1)^2 - 2 y = a(x^2+2x+1) - 2 y = ax^2+2ax+a - 2 Now because the constant term is 0, this means a-2 = 0 ---> a = 2 So because a = 2, we know y = ax^2+2ax + a-2 y = 2x^2+2*2x+2-2 y = 2x^2+4x So the graph of y = 2x^2+4x will have a vertex of (-1,-2)
ooooooooooooh problem i miss typed that its for what values of a and b will the vertex of the graph of y=ax^2 +bx be located at (-1, -2)?
I knew what you meant, so no worries
ooooooooooooooh thank you
you're welcome
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