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".
+33615 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?
