How to Graph a Parabola

Find the vertex, axis of symmetry, and intercepts, then connect them with a smooth U-shaped curve. For y = x² − 4, the vertex is (0, −4) and the roots sit at x = ±2.

You can check your sketch with the parabola graphing calculator after you place the key points by hand. Matching your sketch to a live plot is a fast way to catch an off-by-one vertex or a missed root.

What is a parabola (y = ax² + bx + c)?

A parabola is the graph of a quadratic. In y = ax² + bx + c, a, b, and c are constants and a ≠ 0.

The sign of a sets the direction: a > 0 opens upward; a < 0 opens downward. The size of |a| stretches or flattens the curve. The constants b and c shift the graph left-right and up-down through the vertex and intercepts.

Every upward or downward parabola has one vertex (the turning point) and a vertical axis of symmetry through that vertex.

How do you find the vertex and axis of symmetry?

The x-coordinate of the vertex is:

x = −b / (2a)

Substitute that x into the equation to get the y-coordinate. The axis of symmetry is the vertical line x = −b/(2a).

For y = x² − 4, rewrite as y = 1·x² + 0·x − 4. Then a = 1, b = 0, c = −4:

• x = −0 / (2·1) = 0
• y = 0² − 4 = −4
• vertex (0, −4)
• axis x = 0

If b were nonzero, the vertex would sit off the y-axis, but the same formula still applies.

How do you find intercepts and roots?

Y-intercept: set x = 0. Then y = c, so the point is (0, c). For y = x² − 4, that is (0, −4), which matches the vertex in this special case.

X-intercepts (roots): set y = 0 and solve ax² + bx + c = 0.

For y = x² − 4:

x² − 4 = 0
(x − 2)(x + 2) = 0
x = 2 or x = −2

So the roots are x = ±2, and the points on the graph are (2, 0) and (−2, 0).

When the quadratic does not factor cleanly, use the quadratic formula. When the discriminant is negative, there are no real x-intercepts; the curve still has a vertex and a y-intercept.

Worked example: graph y = x² − 4

1. Vertex: (0, −4). Axis: x = 0.
2. Roots: (±2, 0) from (x − 2)(x + 2) = 0.
3. Extra points (use symmetry across x = 0):
   • x = 1 → y = 1 − 4 = −3 → (1, −3) and (−1, −3)
   • x = 3 → y = 9 − 4 = 5 → (3, 5) and (−3, 5)
4. Plot the vertex, roots, and extras. Draw a smooth curve through them. Do not connect the points with straight segments; the sides should bend gently.

The finished sketch opens upward from (0, −4), crosses the x-axis at −2 and 2, and climbs through (±3, 5).

Common mistakes when graphing a parabola

• Wrong sign in the vertex formula. Remember x = −b/(2a), not +b/(2a). With b = 0 this error is hidden; with b ≠ 0 it shifts the whole graph.
• Forgetting the axis of symmetry. Once you have one side point, the mirror point across the axis is free and keeps the sketch balanced.
• Joining points with straight lines. A parabola is curved. Straight segments between plotted points make a polygonal shape that is not the graph.
• Mixing up roots and the vertex. Roots are where y = 0. The vertex is the turning point; for y = x² − 4 they are different points except that the y-intercept and vertex happen to coincide.

If your curve does not pass through (±2, 0) for this equation, recheck the factoring before redrawing.

FAQ

How do you graph a parabola from y = ax² + bx + c?

Find the vertex with x = −b/(2a), compute y at that x, then find the axis of symmetry, intercepts, and a few extra points. Plot those points and connect them with a smooth curve. For y = x² − 4, the vertex is (0, −4) and the roots are at x = ±2.

How do you find the vertex of a parabola?

Use x = −b/(2a), then substitute that x into the equation to get y. For y = x² − 4, a = 1 and b = 0, so x = 0 and y = −4. The vertex is (0, −4).

How do you find the roots of y = x² − 4?

Set y = 0: x² − 4 = 0. Factor as (x − 2)(x + 2) = 0, so x = 2 or x = −2. The x-intercepts on the graph are (2, 0) and (−2, 0).

Does a positive a open upward?

Yes. When a > 0 the parabola opens upward (a U shape). When a < 0 it opens downward. In y = x² − 4, a = 1, so the curve opens upward from the vertex (0, −4).

What is the axis of symmetry?

It is the vertical line through the vertex, x = −b/(2a). Points on either side of that line are mirror images. For y = x² − 4 the axis is x = 0 (the y-axis).