# Finding Equation of a Parabola?

• Jul 30th 2012, 09:09 AM
woahitzme
Finding Equation of a Parabola?
Determine the equation of the parabola:

Given two points at (-4,9) and (12,9). I figured out the axis of symmetry is x=4. Also given the y-intercept, which is (0,-3). Figure out the equation?

Thanks for the help! (Itwasntme)
• Jul 30th 2012, 09:30 AM
richard1234
Re: Finding Equation of a Parabola?
Write the parabola in the form $y = a(x-b)^2 + c$ where $a,b,c$ are constants and $a \neq 0$.

It is symmetric about $x = 4$, so letting $b = 4$ yields our desired parabola.

Now our parabola is in the form $y = a(x-4)^2 - c$. Subsitutute the given values of (x,y) to obtain a system of equations in terms of $a$ and $c$.
• Jul 30th 2012, 09:32 AM
Wilmer
Re: Finding Equation of a Parabola?
Quote:

Originally Posted by woahitzme
Determine the equation of the parabola:
Given two points at (-4,9) and (12,9). I figured out the axis of symmetry is x=4. Also given the y-intercept, which is (0,-3). Figure out the equation?

Get a sheet of graph paper and plot those 2 points. Then come back and post your question correctly.
If you can't tell right away that y-intercept is (0,9), then you need classroom help.
• Jul 30th 2012, 09:34 AM
richard1234
Re: Finding Equation of a Parabola?
Quote:

Originally Posted by Wilmer
Get a sheet of graph paper and plot those 2 points. Then come back and post your question correctly.
If you can't tell right away that y-intercept is (0,9), then you need classroom help.

It's a parabola, not a line.
• Jul 30th 2012, 09:50 AM
Prove It
Re: Finding Equation of a Parabola?
Quote:

Originally Posted by woahitzme
Determine the equation of the parabola:

Given two points at (-4,9) and (12,9). I figured out the axis of symmetry is x=4. Also given the y-intercept, which is (0,-3). Figure out the equation?

Thanks for the help! (Itwasntme)

The quadratic has the equation \displaystyle \begin{align*} y = a\,x^2 + b\,x + c \end{align*}. Substituting the y-intercept \displaystyle \begin{align*} (0, -3) \end{align*} yields \displaystyle \begin{align*} c = -3 \end{align*}, giving \displaystyle \begin{align*} y = a\,x^2 + b\,x - 3 \end{align*}.

Substitute each of the other points and solve the two equations simultaneously for a and b.
• Jul 30th 2012, 09:55 AM
Wilmer
Re: Finding Equation of a Parabola?
Quote:

Originally Posted by richard1234
It's a parabola, not a line.

Yikes...guess I'm the one who needs classroom help !!
• Jul 30th 2012, 10:02 AM
woahitzme
Re: Finding Equation of a Parabola?
Quote:

Originally Posted by richard1234
Write the parabola in the form $y = a(x-b)^2 + c$ where $a,b,c$ are constants and $a \neq 0$.

It is symmetric about $x = 4$, so letting $b = 4$ yields our desired parabola.

Now our parabola is in the form $y = a(x-4)^2 - c$. Subsitutute the given values of (x,y) to obtain a system of equations in terms of $a$ and $c$.

I plugged in (-4,9) as well as the y-intercept which was -3 for c. I got $9= a(8)^2 - 3$ and ended up getting $a= 3/16$.

The book answer is that the equation is $1/4 x^2 - 2x -3$.
• Jul 30th 2012, 10:11 AM
woahitzme
Re: Finding Equation of a Parabola?
Nevermind, I got the answer, thank you guys!