Determining the equation of a parabola

**Lepzed** · July 31st 2011, 03:56 AM

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 $(x_{top}, y_{top})$ is given: $y = a(x - x_{top})^2 + y_{top}$ .

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

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

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.

**Stroodle** · July 31st 2011, 04:20 AM

You disregarded the translation. You need to expand $y=-\frac{1}{8}(x-3)^2$

**Lepzed** · July 31st 2011, 04:33 AM

Ah, that makes sense

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

Thanks

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