Find the quadratic function whose vertex is (-7,6) and x-intercept is 4

Answer 1

y = a(x+7)^2 +6

plug in the x intercept (4,0) x=4, y=0 to solve for "a", the coefficient of the x^2 term

0 =a(4+7)^2 +6 = 121a +6

121a =-6

a = -6/121

y = f(x) = (-6/121)(x+7)^2 + 6 is one quadratic function with vertex (-7,6) and x intercept = 4

It's a downward opening parabola with maximum y = 6. It also has a 2nd x intercept = -18

another quadratic function with the same vertex and x intercept is half of a rightward opening parabola

x = a(y-6)^2 -7

4 =a(0-6)^2 - 7 = 36a -7

36a = 11

a = 11/36

x = f(y) = (11/36)(y-6)^2 - 7, but limit y to y<6

that is the bottom half of the rightward opening parabola, which makes it a function. Minimum x = -7

the full parabola is not a function, although it is a quadratic relation. the full parabola fails the vertical line test for a function, as a vertical line cuts the parabola twice.

You can't use the top half of the parabola as it has no x intercept