How to Graph a Quadratic Function (Parabola) Step by Step

To graph a quadratic $y = ax^2 + bx + c$, find five things: which way it opens (the sign of $a$), the vertex, the axis of symmetry, the y-intercept and the x-intercepts (roots). Plot them, then draw a smooth, symmetric U-shaped curve called a parabola.

Step 1: Which way does it open?

• $a > 0$: opens up like a smile, with a minimum point.
• $a < 0$: opens down like a frown, with a maximum point.

The bigger $|a|$ is, the narrower the parabola.

Step 2: Find the vertex and axis of symmetry

Substitute that $x$ back into the equation to get the $y$-coordinate of the vertex. The vertical line through the vertex is the axis of symmetry.

Step 3: Find the intercepts

• y-intercept: set $x = 0$, which gives $y = c$.
• x-intercepts: solve $ax^2 + bx + c = 0$ by factoring or the quadratic formula. If $b^2 - 4ac < 0$, the graph does not cross the x-axis.

Worked example 1: $y = x^2 - 4x + 3$

1. $a = 1 > 0$, so it opens up.
2. Vertex: $x = -\frac{-4}{2} = 2$, and $y = 4 - 8 + 3 = -1$, so the vertex is $(2, -1)$. In vertex form, $y = (x - 2)^2 - 1$.
3. y-intercept: $(0, 3)$.
4. x-intercepts: $x^2 - 4x + 3 = (x - 1)(x - 3) = 0$, so $x = 1$ and $x = 3$.

Plot $(1, 0)$, $(2, -1)$, $(3, 0)$ and $(0, 3)$. By symmetry about $x = 2$, $(4, 3)$ is also on the curve.

Worked example 2: $y = -x^2 + 2x + 3$

1. $a = -1 < 0$, so it opens down.
2. Vertex: $x = -\frac{2}{-2} = 1$ and $y = -1 + 2 + 3 = 4$, so the maximum is at $(1, 4)$.
3. y-intercept: $(0, 3)$.
4. Roots: $-x^2 + 2x + 3 = 0 \Rightarrow x = -1$ or $x = 3$.

A table of values

If you are unsure, make a table: choose a few $x$-values either side of the vertex, work out $y$ for each, and plot the points.

Common mistakes

• Drawing a V shape instead of a smooth curve.
• Getting the sign of the vertex's $x$-coordinate wrong: it is $-\frac{b}{2a}$.
• Forgetting that the curve keeps going up (or down) forever.

Practice questions

1. Does $y = 2x^2 - 8$ open up or down, and where is its vertex?
2. Find the roots of $y = x^2 - 9$.
3. Find the vertex of $y = x^2 + 6x + 5$.

Answers: 1) Up, vertex $(0, -8)$ 2) $x = \pm 3$ 3) $(-3, -4)$

See the parabola drawn with its roots and vertex in the quadratic equation solver with graph, convert to vertex form with the vertex form calculator, or plot any curve in the graphing calculator. For solving methods, read three ways to solve a quadratic equation.

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