Find the equation of a parabola

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In summary, the equation of a parabola with the following characteristics is y = a(x-p)^2 + q. Substituting in the y-intercept of -6 and q from the equation, this gives us y = 16a + 6q. When solving for q, the y-intercept gives us 0= 0, confirming that the parabola has a turning point at (-4, 8).
  • #1
ChieF.
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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
 
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  • #2
ChieF. said:
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.
 
  • #3
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
 
  • #4
thanks for the help
 

FAQ: Find the equation of a parabola

1. What is a parabola?

A parabola is a symmetrical, U-shaped curve that is formed by the graph of a quadratic equation. It is a conic section and can be found in various real-world applications, such as bridges, satellite dishes, and projectile motion.

2. How do you find the equation of a parabola?

The general equation of a parabola is y = ax^2 + bx + c, where a, b, and c are constants. To find the equation, you will need to have three points on the parabola or the vertex and one other point. Then, you can plug in the values of those points into the general equation and solve for a, b, and c.

3. What is the vertex form of a parabola?

The vertex form of a parabola is y = a(x-h)^2 + k, where (h,k) is the vertex of the parabola and a is a constant that determines the shape and direction of the parabola. This form is useful for graphing and finding the vertex of a parabola.

4. Can a parabola have a negative coefficient?

Yes, a parabola can have a negative coefficient. The coefficient a in the general equation (y = ax^2 + bx + c) determines the direction and shape of the parabola. If a is negative, the parabola will open downwards, and if a is positive, the parabola will open upwards.

5. How are parabolas used in real life?

Parabolas have many real-world applications. For example, they are used in architecture to design arches and bridges that can withstand weight and stress. They are also used in satellite dishes to reflect signals to a focal point. In physics, parabolas are used to model projectile motion, such as the trajectory of a thrown ball.

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