Finding Equation of a Parabola?

**woahitzme** · July 30th 2012, 08:09 AM

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!

**richard1234** · July 30th 2012, 08:30 AM

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 $c$ .

**Wilmer** · July 30th 2012, 08:32 AM

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.

**richard1234** · July 30th 2012, 08:34 AM

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.

**Prove It** · July 30th 2012, 08:50 AM

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!

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.

**Wilmer** · July 30th 2012, 08:55 AM

Originally Posted by richard1234

It's a parabola, not a line.

Yikes...guess I'm the one who needs classroom help !!

**woahitzme** · July 30th 2012, 09:02 AM

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 $1/4 x^2 - 2x -3$ .

**woahitzme** · July 30th 2012, 09:11 AM

Nevermind, I got the answer, thank you guys!

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