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Sketching a parabola

The graph of a quadratic function $y=ax^2+bx+c$ (where $a\ne 0$) is called a parabola. The sign of $a$ determines whether the parabola opens upward ($a>0$) or opens downward ($a<0$). The “turning point”1) of a parabola is called the vertex.

In this article, we will describe how to sketch any parabola.


Let's say we want to sketch the graph of $y=2x^2-4x-1$.

Get the basic information

The basic information about a parabola is its vertex and whether it opens upward or downward.


Recall that the $x$-coordinate of the vertex of the parabola $y=ax^2+bx+c$ is given by \[x=-\frac{b}{2a}\]

In this case, $x=-\frac{-4}{4}=1$. To find the $y$-coordinate, we plug in $x=1$ into the equation: $y=2-4-1=-3$. Thus the vertex is \[(1,-3)\]

Opens upward or downward

Since in this case $a=2>0$, the parabola opens upward.

Start drawing, add intercepts

Starting from the location of the vertex, draw an open-upward parabola.

At this point, we need to decide how to draw the graph (specifically where will it cross the $y$-axis).


To sketch the graph accurately, we need to know whether the graph will cross over the $y$-axis above or below the $x$-axis. That is, we need to first find the $y$-intercept. This is easy, we just plug in $x=0$. In this case, the $y$-intercept is $y=-1$ which means that the parabola will cross the $y$-axis below the $x$-axis.

From this sketch, we also know that the function has two roots ($x$-intercepts): one positive and the other one negative.


We can also optionally find the $x$-intercepts. If the function cannot be factored easily, we use the quadratic formula.

The solutions of $ax^2+bx+c=0$ are given by \[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]

In this example, the $x$-intercepts are $\displaystyle x=\frac{2-\sqrt{6}}{2}, \frac{2+\sqrt{6}}{2}$. Hence, the complete sketch is

See also

The Lowest point if the parabola opens upward or highest point if it opens downward
Last modified: 2017/01/05 08:30