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The graph of a quadratic function is a parabola, which is a "u"-shaped curve. In this article, we review how to graph quadratic functions.
The graph of a quadratic function is a parabola, which is a "u"-shaped curve:
Looking for an introduction to parabolas? Check out this video.

Example 1: Vertex form

Graph the equation.
$y=-2\left(x+5{\right)}^{2}+4$
This equation is in vertex form.
$y=a\left(x-h{\right)}^{2}+k$
This form reveals the vertex, $\left(h,k\right)$, which in our case is $\left(-5,4\right)$.
It also reveals whether the parabola opens up or down. Since $a=-2$, the parabola opens downward.
This is enough to start sketching the graph.
To finish our graph, we need to find another point on the curve.
Let's plug $x=-4$ into the equation.
$\begin{array}{rl}y& =-2\left(-4+5{\right)}^{2}+4\\ \\ & =-2\left(1{\right)}^{2}+4\\ \\ & =-2+4\\ \\ & =2\end{array}$
Therefore, another point on the parabola is $\left(-4,2\right)$.
Want another example? Check out this video.

Example: Non-vertex form

Graph the function.
$g\left(x\right)={x}^{2}-x-6$
First, let's find the zeros of the function—that is, let's figure out where this graph $y=g\left(x\right)$ intersects the $x$-axis.
$\begin{array}{rl}g\left(x\right)& ={x}^{2}-x-6\\ \\ 0& ={x}^{2}-x-6\\ \\ 0& =\left(x-3\right)\left(x+2\right)\end{array}$
So our solutions are $x=3$ and $x=-2$, which means the points $\left(-2,0\right)$ and $\left(3,0\right)$ are where the parabola intersects the $x$-axis.
To draw the rest of the parabola, it would help to find the vertex.
Parabolas are symmetric, so we can find the $x$-coordinate of the vertex by averaging the $x$-intercepts.
With the $x$-coordinate figured out, we can solve for $y$ by substituting into our original equation.
$\begin{array}{rl}g\left(0.5\right)& =\left(0.5{\right)}^{2}-\left(0.5\right)-6\\ \\ & =0.25-0.5-6\\ \\ & =-6.25\end{array}$
Our vertex is at $\left(0.5,-6.25\right)$, and our final graph looks like this:
Want another example? Check out this video.

Practice

Problem 1
Graph the equation.
$y=2\left(x+1\right)\left(x-1\right)$

Want more practice graphing quadratics? Check out these exercises:

Want to join the conversation?

• How even do you guys DO that?!
• Do some more Khan Academy videos and exercises. You'll get the hang of it sooner of later if you try! Good luck!
• how to find the vertex
• One way is to complete the square and put in vertex form, Another way is to use -b/2a as the x coordinate and then use that to solve for y.
• We only use two point here to graph a parabola but I've heard that it takes three points to define one. Is it true that an infinite number of parabolas can be drawn through just two distinct points? on the coordinate plane?
• If you know the vertex and another point, then you also know the reflexive point, so you have 3 points. So two points work as long as one of them is the vertex.
• On paper, how would I find the curve to nicely draw it out? Would I have to plot several points to graph it out by hand or is there another rule?
• Yes, the more points you plot, the more accurate you can draw the curve. Some turn the graph paper sideways and draw on one side of the vertex, rotate it 180 degrees and draw on the other side of the vertex.
• Question? WHere did the word quadratics come from?
• It comes from the Latin word quadratum, which means square.
• Do i need to learn these, or will it come back to me if i skip parbolas?
• Yes, don't skip it. Try to understand it, and it will save you a lot of frustration later on.
• On example no.1. How did he came up with x=-4 to plug into the equation?

And also in Practise problem no.2, how did he came up with x=0 to plug into the equation?
• On Example 1, the vertex was at (-5,4). Sal moved over 1 unit to pick x=-4. I believe he did this to demonstrate the impact of the -2 in front. It reduced the y-value of 4 by 2 units.

On practice problem 2, I'm sure he picked x=0 to find the y-intercept of the equation.
• why plugged x=−4x,when you can do y=0
• Sal has the equation: y = -2(x+5)^2+4. This is vertex form. So, you graph the vertex and then find points to the left and right of the vertex.

When Y is isolated already and the equation is in vertex form, it is easier to pick values of X and calculate Y especially when you have a 2nd degree (quadratic) equation or higher degree. The math is usually simpler.

If the equation had been in standard form: y=Ax^2+Bx+C, then it is quite common to find the x-intercepts 1st (set y=0 and calculate X). Though, you can also get the vertex pretty easily from this form and work outward from there.

The benefit of having the vertex is that you know the highest / lowest point in the graph and you know the graph will be symmetrical as it moves away from the vertex.