The vertex form for parabolas is convenient for graphing, but often times you'll be given the equation in the usual quadratic form instead.

*y* = *ax*^{2} + *bx* + *c*

Can we deduce a way to find the vertex of a parabola from its quadratic form equation? Would we even ask the question if the answer was "No?"

Let's start with the vertex form function, and expand it to its quadratic form.

*y* = *a*(*x* – *h*)^{2} + *k*
*y* = *ax*^{2} + (-2*ah*)*x* + (*ah*^{2} + *k*)

Remember? We did this in the last section. It wasn't *that* long ago. Anyway, if you compare the second function above to the standard quadratic function, you'll notice that -2*ah* = *b*. And hey—if we're given a quadratic function, we know what *a* and *b* are. If only we knew what -2 equaled, we could solve for *h*. Wouldn't that be great? Oh well, let's solve for *h* anyway, and pretend we do know. Your teacher won't notice, right?

While we could go ahead and solve *c* = *ah*^{2} + *k* for *k*, there is a much, much better way. *h* is a value of *x* just like any other, so we can just plug *h* into the function and solve. That means that the vertex equals (*h*, *f*(*h*)), or . This way will be much more convenient to use. Smarter, not harder, remember?

Graph the function *y* = 4*x*^{2} + 6*x* – 2.

Let's start off by looking for the vertex. Using our handy-dandy new formula, we find

.

Next we plug *h* in for *x* to find *y*. Golly, that's a lot of variables.

So the vertex is .

The *y*-intercept is easy to find, being equal to 4(0)^{2} + 6(0) – 2 = –2, so (0, -2).

Next we'll check to see if any *x*-intercepts exist. The discriminate is b^{2} – 4*ac* = 6^{2} – 4(4)(-2) = 36 + 32 = 68. We have two real number roots, and so two *x*-intercepts. Quadratically formulating a solution, we get

Notice that we already know what goes under the square root, because we calculated the discriminate.

and are our *x*-intercepts, which are equal to about (0.281, 0) and (-1.781, 0).

These are enough points for us to graph the function, so let's do that.

As we can see, once we have the vertex graphing a parabola in the quadratic form is very similar to doing it in the vertex form.

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