Find the equation of a parabola

[ChieF.]

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

 

[HallsofIvy]

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

Well, of course, solving for q using the y intercept and then putting the y-intercept into the equation will give you 0= 0! What did you expect?

You haven't used all of the information. The fact that "range y<= 8" tells you that the vertex is at (-4, 8). That tells you both p and q. After you know p and q, the fact that the parabola passes through (0, -6) will give you a, which must be negative as the parabola opens downward.

 

[yeongil]

I'm assuming that "turning point" = "vertex"...

If "range Y <= 8" then doesn't that imply that the y-coordinate of the vertex is 8? Can you take it from there?EDIT: Beaten to the punch by HallsofIvy! ;)

01

 

[ChieF.]

thanks for the help