Originally Posted by
yeongil If your graph looks like a "squiggly line," then you got your signs mixed up somewhere.
To graph this, you could rewrite in vertex form:
$\displaystyle f(x) = a(x - h)^2 + k$
So,
$\displaystyle f(x) = 15 - x - x^2$
$\displaystyle f(x) = -x^2 - x + 15$
$\displaystyle f(x) = -(x^2 + x) + 15$
$\displaystyle f(x) = -\left(x^2 + x + \frac{1}{4}\right) + 15 + \frac{1}{4}$
$\displaystyle f(x) = -\left(x + \frac{1}{2}\right)^2 + \frac{61}{4}$
Since the leading coefficient was negative, the parabola opens downward. The vertex is given as (h, k), so it's $\displaystyle \left(-\frac{1}{2},\;\frac{61}{4}\right)$. You can also find the x- and y- intercepts to get additional points.
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