I can't seem to figure this problem out, there doesn't seem to be enough information.

Find the standard equation of the parabola that has a vertical axis that has $x$-intercepts $-5$ and $3$ and has a lowest point with a $y$-coordinate of $-7$.

I could either do $0 = a(-5 -h)^2 -7$ or $0 = a(3 - h)^2 - 7$ but either way it seems like I don't have enough information. I guess I could just find $a$ in terms of $h$ or viceversa but I feel like there is an easier solution. Any advice?


Use the general equation $y=a(x-h)^2+k$, where $(h,k)$ are the coordinates of the vertex.

You already know that the $y$-coordinate of the vertex is $k=-7$. The $x$-coordinate of the vertex, $h$, is given to you indirectly: it is by symmetry the midpoint of the $x$-intercepts. So, find the value of $h$ using this and then substitute the known values of $h$ and $k$ into the general equation.

This still leaves you the unknown $a$. But, you know that the point $(3,0)$ is on the parabola, and you can substitute $x=3$ and $y=0$ into your equation and solve for $a$.

  • $\begingroup$ Oh yeah I forgot about finding a midpoint. Thanks :) $\endgroup$ – Tyler Jarjoura Oct 2 '12 at 19:40

If the parabola has two $x$-intercepts $a$ and $b$, it is given by $y=c(x-a)(x-b)$ for some constant $c$, with $c\ne0$ of course. (If you're not sure why, try substituting $a$ or $b$ in the equation.)

Since your parabola has vertical axis of symmetry, its minimum occurs when $x = \frac{a+b}{2}$, that is, halfway between the two $x$-intercepts.

So we have $$y=c(x-(-5))(x-3)\\ y=c(x+5)(x-3)$$ for some yet to be determined value $c$.

The parabola has a minimum of $-7$ when $x=\frac{5-(-3)}{2}=1$, so you know that $$-7=c(1+5)(1-3)\\ -7=c\cdot6\cdot(-2)\\ \frac{7}{12}=c$$

So your parabola is given by $$y=\frac{7}{12}(x+5)(x-3)$$




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