write a quadratic function whose graph passes through (4,-7) and has a vertex of (1,6)
REmember that
The equation of a vertical parabola in vertex form is equal to
y=a(x-h)^2+k
where
(h,k) is the vertex
a is the leading coefficient
substitute the given values
(h,k)=(1,6)
y=a(x-1)^2+6
Find the value of a
we have the point (4,-7)
so
For x=4
y=-7
substitute
-7=a(4-1)^2+6
-7=a(9)+6
solve for a
9a=-7-6
9a=-13
a=-13/9
therefore
y=(-13/9)(x-1)^2+6