Graphs of Second Degree Polynomials

How the leading coefficient and the discriminant shape a parabola

Overview

A second-degree polynomial — a quadratic function \( f(x) = ax^2 + bx + c \) with \( a \ne 0 \) — has a graph called a parabola. Two algebraic quantities together determine its shape:

The Six Combinations

Combining the two signs gives six possible parabola behaviors. Each graph below is one of these cases.

Leading coefficient Discriminant Orientation x-intercepts
\( a > 0 \)\( D > 0 \)Upward2
\( a > 0 \)\( D = 0 \)Upward1
\( a > 0 \)\( D < 0 \)Upward0
\( a < 0 \)\( D > 0 \)Downward2
\( a < 0 \)\( D = 0 \)Downward1
\( a < 0 \)\( D < 0 \)Downward0

Worked Examples

Example 1 — \(f(x) = x^2\) (opens upward, one x-intercept)

Question: Why does the parabola \( f(x) = x^2 \) open upward and touch the x-axis at exactly one point?

Graph of f(x) = x² showing an upward-opening parabola touching the x-axis at the origin
Figure 1 — \( f(x) = x^2 \).
View answer

Orientation. The leading coefficient is \( a = 1 > 0 \), so the parabola opens upward.

Number of x-intercepts. Solve \( x^2 = 0 \): the double zero is \( x = 0 \). Algebraically, the discriminant is:

\[ D = b^2 - 4ac = 0^2 - 4(1)(0) = 0 \]

Since \( D = 0 \), there is exactly one real zero (double root). The vertex is at \( (0, 0) \) — right on the x-axis — so the parabola touches at exactly one point.

Example 2 — \(f(x) = x^2 - 2x\) (opens upward, two x-intercepts)

Question: Why does \( f(x) = x^2 - 2x \) intersect the x-axis at two distinct points?

Graph of f(x) = x² − 2x showing an upward-opening parabola with x-intercepts at x = 0 and x = 2
Figure 2 — \( f(x) = x^2 - 2x \).
View answer

Factor to find the zeros.

\[ x^2 - 2x = x(x - 2) = 0 \]

So the zeros are \( x = 0 \) and \( x = 2 \).

Confirm with the discriminant. Here \( a = 1, b = -2, c = 0 \):

\[ D = (-2)^2 - 4(1)(0) = 4 > 0 \]

Since \( D > 0 \), there are two distinct real zeros, and the parabola crosses the x-axis twice.

Example 3 — \(f(x) = -2x^2 - 3x + 2\) (opens downward, two x-intercepts)

Question: Why does \( f(x) = -2x^2 - 3x + 2 \) open downward, and where does it cross the x-axis?

Graph of f(x) = −2x² − 3x + 2 showing a downward-opening parabola with x-intercepts at x = −2 and x = 0.5
Figure 3 — \( f(x) = -2x^2 - 3x + 2 \).
View answer

Orientation. The leading coefficient is \( a = -2 < 0 \), so the parabola opens downward.

Number of x-intercepts. Here \( a = -2, b = -3, c = 2 \). Discriminant:

\[ D = (-3)^2 - 4(-2)(2) = 9 + 16 = 25 > 0 \]

So there are two real zeros. Using the quadratic formula:

\[ x = \frac{3 \pm \sqrt{25}}{-4} = \frac{3 \pm 5}{-4} \]

Giving \( x = -2 \) or \( x = \tfrac{1}{2} \). Both x-intercepts are visible in the figure.

Example 4 — \(f(x) = x^2 + 3x + 3\) (opens upward, no x-intercepts)

Question: Why does the graph of \( f(x) = x^2 + 3x + 3 \) have no x-intercepts?

Graph of f(x) = x² + 3x + 3 showing an upward-opening parabola entirely above the x-axis
Figure 4 — \( f(x) = x^2 + 3x + 3 \).
View answer

Orientation. \( a = 1 > 0 \), so the parabola opens upward — it has a minimum somewhere on its axis of symmetry.

Number of x-intercepts. The discriminant is:

\[ D = b^2 - 4ac = 3^2 - 4(1)(3) = 9 - 12 = -3 < 0 \]

Since \( D < 0 \), there are no real zeros. The minimum of the parabola lies above the x-axis, so the graph never intersects it.

Practice Problems

For each quadratic below, (1) state whether the parabola opens upward or downward, (2) compute the discriminant, and (3) determine the number of x-intercepts.

Practice 1 — \( f(x) = x^2 - 4x + 4 \)

Answer

\( a = 1 > 0 \) — opens upward.

\( D = (-4)^2 - 4(1)(4) = 16 - 16 = 0 \) — one x-intercept.

Indeed \( f(x) = (x - 2)^2 \), so the vertex \( (2, 0) \) is on the x-axis.

Practice 2 — \( f(x) = -x^2 + 2x - 3 \)

Answer

\( a = -1 < 0 \) — opens downward.

\( D = 2^2 - 4(-1)(-3) = 4 - 12 = -8 < 0 \) — no x-intercepts.

The parabola lies entirely below the x-axis.

Practice 3 — \( f(x) = 2x^2 - 5x + 2 \)

Answer

\( a = 2 > 0 \) — opens upward.

\( D = 25 - 16 = 9 > 0 \) — two x-intercepts.

Factor: \( 2x^2 - 5x + 2 = (2x - 1)(x - 2) \), so x-intercepts are at \( x = 1/2 \) and \( x = 2 \).

Frequently Asked Questions

How can you tell whether a parabola opens upward or downward?

Look at the leading coefficient \(a\) in \(f(x) = ax^2 + bx + c\). If \(a > 0\), the parabola opens upward; if \(a < 0\), it opens downward. The magnitude of \(a\) controls width but not orientation.

What does the discriminant tell you about a quadratic's graph?

The discriminant \(D = b^2 - 4ac\) determines the number of x-intercepts. \(D > 0\) → two distinct real zeros (two x-intercepts). \(D = 0\) → one double zero (one x-intercept, at the vertex). \(D < 0\) → no real zeros (no x-intercepts).

What is the vertex of a parabola, and how do you find it?

The vertex is the highest (if \(a < 0\)) or lowest (if \(a > 0\)) point. Its x-coordinate is \(x = -b/(2a)\); its y-coordinate is \(f(-b/(2a))\).

Can a quadratic have no x-intercepts?

Yes — when \(D = b^2 - 4ac < 0\). For example, \(f(x) = x^2 + 3x + 3\) has \(D = -3 < 0\), so its graph never crosses the x-axis.

More References on Polynomial Functions

Last updated: October 17, 2026