SOLUTION: what is the equation of a parabola graph? The parabola opens downward. The vertex is (0,2). It also has points (-2,-2) and (2,-2)

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: what is the equation of a parabola graph? The parabola opens downward. The vertex is (0,2). It also has points (-2,-2) and (2,-2)      Log On


   



Question 206447: what is the equation of a parabola graph?
The parabola opens downward. The vertex is (0,2). It also has points (-2,-2) and (2,-2)

Found 2 solutions by stanbon, mickclns:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
what is the equation of a parabola graph?
The parabola opens downward. The vertex is (0,2). It also has points (-2,-2) and (2,-2)
--------------------------
Form: y = a(x-h)^2+k
Vertex: h = 0, k = 2
Ao,y = ax^2 +2
---
Using (-2,-2) and (2,-2) you get
-2 = 4a +2
a = -1
================
Equation:
y = -x^2+2
=================
Cheers,
Stan H.

Answer by mickclns(59) About Me  (Show Source):
You can put this solution on YOUR website!
There are several forms in which the equation of a parabola can be written. One of them is vertex form, because the vertex can be read directly off of it. the form is:
+y+=+a%28x-h%29%5E2+%2B+k+ where (h,k) is the vertex, occasionally written as +y+-+k+=+a%28x-h%29%5E2+. If the parabola opens up then a is a positive number, and if the parabola opens down then a is a negative number.
 
If the parabola opens right (a+) or left a-), then the equation would be +x+=+a%28y-k%29%5E2+%2Bh+ where (h,k) is the vertex, or x+-+h+=+%28y+-+k%29%5E2+. (note the y part is squared and the x part is not)
 
In the case of yours, then you have +y+=+a%28x-0%29%5E2+%2B+2+ and a is negative. So, how do you find a? Well, you talk about an equation having points. That means if you plug the x and the y coordinates of the point into x and y, respectively, the equation you should come out with a true statement/equation. So, let's take one of the other points (the vertex won't help you find a (Why?) ), how about (2,-2) ... when you plug 2 in for x and -2 in for y, you have:
+-2+=+a%282-0%29%5E2+%2B+2+
+-4+=+4a+%2B+2+
+a=-1
So an equation for your parabola is +y+=+-1%28x-0%29%5E2+%2B+2+ -- of course the 1 in the -1 and the -0 in the x-0 could be dropped: +y+=+-x%5E2+%2B+2+.
 
The graph looks like:
+graph%28300%2C300%2C-3%2C3%2C-3%2C3%2C-x%5E2+%2B+2%29+
 
Other forms of the equation are +y+=+ax%5E2+%2B+bx+%2B+c+ and +y+=+a%28x-b%29%28x-c%29+
 
- Mick