# How to figure out the equation of a parabola, when only given the graph?

The equation of the graph is y= 2/4x^2 -4, how can I create this equation by only looking at the graph. What does the a,b,c stand for in the ax^2+bx+c? Example in the y intercept equation, y=mx+b, the b is the y intercept. Thanks.

### 3 Answers

- PuzzlingLv 73 weeks ago
The question specifically says to put the equation in *vertex form*

y = a(x - h)² + k

That's vertex form where (h, k) is the vertex.

In your case, you have a vertex at (0, -4)

y = a(x - 0)² - 4

y = ax² - 4

Now you just need one more point. For example, you could use one of the x-intercepts like (4,0). Plug in that point for (x,y) and solve for a:

0 = a(4²) - 4

4 = 16a

a = 4/16

a = 1/4

So the final equation is:

y = (1/4)x² - 4

- davidLv 73 weeks ago
y= 2/4x^2 -4, <<<< but you give the equation, not the graph

... look at the graph, it has 2 x-intercepts and a minimum point .. actually 1/2 way between the intercepts

use y = ax^2 + bx + c >>>> sub each point separately to create 3 equations .... 3 equations with 3 unknowns, a, b, and c ... solve the simultaneous equations .. this gives a, b, and c and use these to write the equation.

- Jeff AaronLv 73 weeks ago
Look at the x-intercepts, which seems to be 4 and -4, so we have:

y = a(x + 4)(x - 4)

y = a(x^2 - 16)

Now look at the y-intercept, which is at (0, -4), so we have:

-4 = a(0^2 - 16)

-4 = a(0 - 16)

-4 = -16a

a = (-4)/(-16)

a = 1/4

y = (1/4)(x^2 - 16)

y = (1/4)x^2 - 4