Determining the equation of a parabola

• Jul 31st 2011, 03:56 AM
Lepzed
Determining the equation of a parabola
I'm asked to give the equation of a parabola when both the top (maximum?*) and a point on the line are given. The general form of a parabola with top $\displaystyle (x_{top}, y_{top})$ is given: $\displaystyle y = a(x - x_{top})^2 + y_{top}$.

Seems straightforward enough and when try to solve this, for example, for $\displaystyle T = (0, 0), P = (1, 2)$, I just fill in the blanks and solve for a: $\displaystyle 2 = a(1 - 0)^2 + 0 \Leftrightarrow 2 = a * 1 \Leftrightarrow a = 2 \Leftrightarrow y = 2x^2$ this sure enough works out as I would've expected.

But then I'm asked to solve for the following: $\displaystyle T = (3, 0), P = (-1, -2)$. Filling in tells me: $\displaystyle -2 = a(-1 - 3)^2 + 0 \Leftrightarrow -2 = a * 16 \Leftrightarrow a = -\frac{1}{8}$ Leaving me to conclude that $\displaystyle y= -\frac{1}{8}x^2$, which according to my textbook is very wrong, it should be: $\displaystyle y = -\frac{1}{8}x^2 + \frac{3}{4}x - \frac{9}{8}$.

Obviously I'm doing this wrong, so I'm looking for some pointers on how to approach a problem like this.

*As English isn't my native language it's possible I confuse terms, in spite of trying to find a fitting translation. I apologise for that on beforehand and hope you would be able to correct me if I do so.
• Jul 31st 2011, 04:20 AM
Stroodle
Re: Determining the equation of a parabola
You disregarded the translation. You need to expand $\displaystyle y=-\frac{1}{8}(x-3)^2$
• Jul 31st 2011, 04:33 AM
Lepzed
Re: Determining the equation of a parabola
Ah, that makes sense :)

$\displaystyle y = -\frac{1}{8}(x - 3)^2 \Leftrightarrow y = -\frac{1}{8}(x^2 - 6x + 9) \Leftrightarrow y = -\frac{1}{8}x^2 + \frac{3}{4}x - \frac{9}{8}$.

Thanks :)