## How to find the equation of a parabola?

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
mitnord
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### How to find the equation of a parabola?

A parabola passes through (0, -1) and (1,1) and is symmetrical about y-axis. How to find the equation of this parabola? Thanks!!

stapel_eliz
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A parabola passes through (0, -1) and (1,1) and is symmetrical about y-axis. How to find the equation of this parabola? Thanks!!
You know that the general equation of a parabola is y = ax2 + bx + c. You've been given:

. . . . .-1 = a(0)2 + b(0) + c

. . . . .1 = a(1)2 + b(1) + c

From the first equation, you can find the value of c. Plug this into the second equation to find a relationship between a and c.

Since the parabola is symmetric, then the point they gave you that is off the y-axis gives you a third point. What is it?

Plug this third point into the general equation. This will give you a second equation that you can simplify for a relationship between a and b.

Solve the system of equations for the values of a and b. Then re-write the general equation, using the values you've found for a, b, and c.

If you get stuck, please reply showing how far you have gotten in working through the steps. Thank you!

mitnord
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Joined: Mon Feb 23, 2009 11:17 pm
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### Re: How to find the equation of a parabola?

Following your steps I found that:

-1 = c and
-a = -b

What to do next? I was only given the two points: (0, -1) and (1, 1) and that the parabola is symmetrical around y-axis. Is it even possible?

DAiv
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Joined: Tue Dec 16, 2008 7:47 pm
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### Re: How to find the equation of a parabola?

Following your steps I found that:

-1 = c and
-a = -b

What to do next? I was only given the two points: (0, -1) and (1, 1) and that the parabola is symmetrical around y-axis. Is it even possible?
c = -1 is correct, but -a = -b isn't. I think you may have made a mistake when dealing with the sign of c.

From the point (1, 1), x = 1 and y = 1

So, since for a parabola:
y = ax2 + bx + c
(1) = a(1)2 + b(1) + c
1 = a + b + c

And from the previous equation, c = -1, so

1 = a + b + (-1) <-- Need to be careful with the minus sign here
1 = a + b - 1

1 + 1 = a + b - 1 + 1
2 = a + b

Subtract b from both sides:

2 - b = a + b - b
2 - b = a

So, a = 2 - b

As Eliz has said, the key piece of information to find the third point is that the parabola is symmetrical around the y-axis. That is, the vertical axis acts like a mirror, and any points to the right of it will be reflected and appear the same distance to the left of it.

The point (0, -1), where x = 0, lies on the axis of symmetry, that is, it is no distance away from it, so its reflection would also be no distance away and therefore would be in the same place, (0, -1). This doesn't tell us anything new, so we can ignore it.

However, the point (1,1) does not lie on the axis of symmetry, so how far away will its reflection be? And since reflecting horizontally doesn't change anything vertically, the y-coordinate will stay the same.

So, from this information, what will be the coordinates of the reflection of (1,1)? This will be your third point which you can then use as outlined by Eliz above.

DAiv