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Find the equation of a parabola with the following characteristics:

range Y <= 8

x-coordinate of the turning point is -4

y-intercept = -6

I have tried to substitute all the information into y = a(x-p)^2 + q

which gives me y = a(x+4)^2 + q and substituted the y-intercept into the equation and then subbed in q into the equation with the y-intercept subbed in as well but that just came up as 0 = 0

Here is my working out

Sub y-intercept into equation

-6 = a(0 +4)^2 +q

-6 = 16a + q

q = -6 -16a

Sub into equation (0,-6) and q

-6 = a(0+4)^2 - 16a -6

0 = 16a - 16a

0 = 0

range Y <= 8

x-coordinate of the turning point is -4

y-intercept = -6

I have tried to substitute all the information into y = a(x-p)^2 + q

which gives me y = a(x+4)^2 + q and substituted the y-intercept into the equation and then subbed in q into the equation with the y-intercept subbed in as well but that just came up as 0 = 0

Here is my working out

Sub y-intercept into equation

-6 = a(0 +4)^2 +q

-6 = 16a + q

q = -6 -16a

Sub into equation (0,-6) and q

-6 = a(0+4)^2 - 16a -6

0 = 16a - 16a

0 = 0

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