# At a Glance - Quadratic Form Parabolas

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 = ax2 + 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?"

y = a(xh)2 + k y = ax2 + (-2ah)x + (ah2 + 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 -2ah = 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 = ah2 + 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?

### Sample Problem

Graph the function y = 4x2 + 6x – 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 b2 – 4ac = 62 – 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.

#### Example 1

 Graph the following quadratic function.y = -x2 – 2x + 6

#### Example 2

 Graph the following quadratic function.y = x2 + 3x + 5

y = 2x2 + x – 7

y = -x2x – 1

y = 3x2 + 4x + 1