# Making the equation of a parabola

1. Aug 10, 2011

### stonecoldgen

1. The problem statement, all variables and given/known data
So i have a parabola graphed, i need to express y in terms of X. The vertex is 2,10 and the roots are 0,0 and 4,0.

2. Relevant equations
ax^2+bx+c=y

3. The attempt at a solution

I thought of working the quadratic in a reverse process, but it doesn't work. Because one of the roots is at 0,0, i guess it doesnt have either b or c. Because the paraboal is upwards and not downards, i guess X must be negative.

2. Aug 10, 2011

### eumyang

Use the vertex form of the quadratic:
y = a(x - h)^2 + k,
where (h, k) is the vertex and a is the same as the a in y = ax^2 + bx + c. Then plug in either root for x and y to solve for a. Once you get the complete vertex form, change it standard form and you'll see what b and c are.

3. Aug 10, 2011

### stonecoldgen

man im feeling dumb, how am i supposed to know what b and c are?

i got to the point where i reachd 2b+c=9.52 after doing some algebra, with the other root, i have another result, so im also confused by that.

4. Aug 10, 2011

### PeterO

There are three possible equations for a parabola

y = ax2 + bx + c
y = a(x-e)2 + f [we don't usually use e and f for constants]
y = a(x-u)(x-v)

Any of them can be used, along with the co-ordinates of any 3 points on the graph, to generate 3 simultaneous equations in 3 unknowns , which can then be solved.

The second is quite useful if you happen to know the Turning Point - you can evaluate some of the constants by inspection.

The third is particularly useful if you know the roots [x intercepts] as again you can solve some constants by inspection.

5. Aug 10, 2011

### Ray Vickson

If f(x) = a*x^2 + b*x + c is a quadratic with roots r1 and r2, it must be of the form k*(x - r1)*(x - r2), because it must vanish at x = r1 and at x = r2, and have powers of x no higher than 2.

RGV

6. Aug 11, 2011

### eumyang

Not sure how you got that. Let me try again. Start with the vertex form:
y = a(x - h)^2 + k

The vertex is (2, 10), so you plug that in for h and k:
y = a(x - 2)^2 + 10

Notice that there are still 3 variables. But we are given two points: (0, 0) and (4, 0). So plug in either x = 0 & y = 0, or x = 4 & y = 0, into the equation above. Now the only unknown is a. Solve for a.

Take this equation below and now plug in the value for a:
y = a(x - 2)^2 + 10

Rewrite the above into standard form (y = ax^2 + bx + c) by squaring the binomial in the parentheses and simplifying. At the end, the number in front of x will be your b value, and the constant will be your c value. Now try it.

7. Aug 11, 2011

### HallsofIvy

Staff Emeritus
vertex is 2,10 and the roots are 0,0 and 4,0.
Putting x= 2, y= 10, x= 0, y= 0, and x= 4, y= 4 into $y= ax^2+ bx+ c$ gives you three linear equations for a, b, and c.

8. Aug 11, 2011

### PeterO

Perhaps you could show us how you managed that.

9. Aug 15, 2011

### stonecoldgen

well, i guess this was the most useful of the posts, thanks