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Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,-1). Express your answer in the form "ax^2+bx+c".

 Nov 21, 2022
 #1
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Parabola

vertex (2,4), point (1,-1)

Find the equation.

 

Hello Guest!

 

\( y-y_v=a(x-x_v)^2\\ y-4=a(x-2)^2\\ y=a(x-2)^2+4\\ -1=a(1-2)^2+4\\ a=\dfrac{-1-4}{(1-2)^2}\\ \color{blue}a=-5\)

\(y=a(x-2)^2+4\\ y=-5(x-2)^2+4\\ y=-5(x^2-4x+4)+4\\ y=-5x^2+20x-20+4\\ y=ax^2+bx+c\\ \color{blue}y=-5x^2+20x-16\)

laugh  !

 Nov 21, 2022
edited by asinus  Nov 23, 2022

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