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Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (7,7). Express your answer in the form "ax^2+bx+c".

 Nov 2, 2022
 #1
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If \(y=ax^2+bx+c\) then usingthe vertex values we have 

\(4=4a+2b+c\)        (1)

 

Using the point (7,7) we have

\(7=49a+7b+c\)      (2)

 

We need a thid equation.  Because of the symmetry about the vertex, the distance on the x-axis of point (7,7) from the vertex is matched by a point the same distance from th vertex in the opposite x direction.  Since point(7,7) is a distance of +5 in the x-direction from (2,4), we must have a point with y-value 7 at x-value 2 - 5 = -3. ie. the point (-3, 7) is also on the curve.  Hence our third equation becomes:

\(7=9a-3b+c\)        (3)

 

 

There are now three equations in three unknowns.  Can you take it from here?

 Nov 2, 2022
edited by Alan  Nov 2, 2022

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