# Quadratics: Graphing Tips

You can graph a quadratic function by plugging in x and y values and plotting out the parabola, just like you can use an x-y table to graph a line.

But, lines are simple because any two points will give you an entire line, parabolas are more complicated. Some open up, while others open down. Some are wide and others are narrow. They start at all different locations on the coordinate plane. Luckily there are a number of "rules" about quadratics that you can use to figure out generally what the graph of a quadratic will look like.

Let's start with a quadratic in standard form:

$x^2-4x+4=0$

What can we learn from that? **First, we can see if the vertex opens upward (like a happy face) or downward (like a sad face).**

Equations in standard form are $ax^2+bx+c=0$ and if $a>0$, the parabola opens upward (happy) and if $a<0$, the parabola opens downward (sad).

*Example*:

In $x^2-4x-12=0$, $a=1$, which is positive, so this parabola opens upward.

That's not all you can learn from the $a$ term of a parabola. **The $a$ also tells you how wide or narrow the parabola is.** A parabola like this one (where $|a|=1$) is a "normal size." If $|a|>1$, the parabola has a vertical stretch, which means that it is narrower. If $|a|<1$, the parabola has a horizontal stretch, which means that it's wide. The more $|a|$ is greater or less than 1, the narrower or wider the parabola will be.

We can also **find the vertex of the parabola** using the formula $x=\dfrac{-b}{2a}$.

*Example*:

In $x^2-4x+4=0$, the x-coordinate of the vertex is $x=\dfrac{-b}{2a}=\dfrac{4}{2}=2$.

So, the vertex is (2, $y$). Plug in the $x$ to find the $y$ of the vertex:

$\eqalign{x^2-4x-12=y\\2^2-4(2)-12=y\\y=-16}$

So, the vertex is (2,-16).

If we put the equation into intercept form, we can also **find the x-intercepts or solutions to the equation** (graphed as the points where the parabola crosses the x-axis). Remember, that to put the quadratic into intercept form, you factor the trinomial into binomials:

*Example*:

$\eqalign{x^2-4x+4=0\\(x-6)(x+2)=0\\x=6, -2}$

There are two real solutions to this quadratic. The parabola crosses the x-axis at 6 and -2.

So, now we know that the equation opens upwards, that its vertex is (2,-16) and that it crosses the x-axis at 6 and -2. In this case, that's probably enough for us to graph the equation. But, this three bits of information are not enough to graph any quadratic. What else can we learn at a glance from an equation?

Let's look at the **vertex form of the parabola: $y=a(x-h)^2+k$ In this form, $(h,k)$ is the vertex of the parabola**. And, the functions of $a$, described above, still apply. So, even in vertex form, it's pretty easy to eyeball a parabola.

*Example*:

$y=4(x+9)^2+6$

In this case, the vertex of the equation is (-9, 6) and because $|4|>1$, and $4>1$, the parabola opens upward and is quite narrow.

These tips will not help you fully solve all quadratic equations, but they will definitely help you eyeball a graph of a parabola and see if it matches an equation!

#### Practice Problems:

## Quadratics: Graphing Tips

Note whether the parabolas represented by the following equations open

*upward*or*downward*and if they are*narrow*,*wide*, or "*normal*."- $y=x^2+9x-8$
- $y=2x^2+5+3$
- $y=-.5x^2+2x-4$
- $y=\dfrac{1}{3}x^2+5x-2$
- $y=-\dfrac{1}{8}x^2+1x-.5$

Draw a quick sketch of the following equations on a coordinate plane.

- $y=2x^2+6x-4$
- $y=-3x^2+5+3$
- $y=\dfrac{1}{2}x^2+8x-1$
- $y=(x-4)^2+5$
- $y=-2(x+3)^2-1$

Match the following equations to the graphs below:

- $y=-\dfrac{1}{2}x^2+2x+3$
- $y=x^2+2x+1$
- $y=2x^2-2x+4$
- $y=-3x^2+2x-2$

#### Answer Key: