Skip to content
Mom's Basement University

Graphing Quadratic Functions

What You’ll Learn

Section titled “What You’ll Learn”

In this lesson you’ll learn how to graph quadratic functions in standard form and identify their important features: vertex, axis of symmetry, direction of opening, minimum and maximum values, and intercepts.

The Concept

Section titled “The Concept”

A quadratic function has the form:

f(x)=ax2+bx+cf(x) = ax^2 + bx + c

where a ≠ 0.

Key features:

  • The graph is a parabola (U-shaped curve).
  • If a > 0, the parabola opens upward (minimum point).
  • If a < 0, the parabola opens downward (maximum point).
  • Vertex: the turning point of the parabola (lowest or highest point).
  • Axis of symmetry: the vertical line through the vertex, given by x = −b / (2a).

To graph:

  1. Find the vertex using x = −b / (2a), then plug it back to find y.
  2. Find the y-intercept (set x = 0).
  3. Find the x-intercepts (set f(x) = 0 and solve, if they exist).
  4. Plot a few more points and sketch the parabola.

Worked Example

Section titled “Worked Example”

Upward-opening parabola: graph f(x) = x² − 4x − 5.

The graph shows the parabola f(x) = x² − 4x − 5 with all its key features labeled. The gold dot marks the vertex at (2, −9), the lowest point on the curve. Since a = 1 > 0 the parabola opens upward, so the vertex is the minimum of the function. That means f(x) never goes below −9, making the range y ≥ −9. The dashed line at x = 2 is the axis of symmetry. The green dots mark the y-intercept at (0, −5) and the two x-intercepts at (−1, 0) and (5, 0).

1. Find the vertex

Here a = 1, b = −4, c = −5. The x-coordinate of the vertex is:

x=−b2a=−−42(1)=2x = -\frac{b}{2a} = -\frac{-4}{2(1)} = 2

Plug x = 2 back in: f(2) = (2)² − 4(2) − 5 = 4 − 8 − 5 = −9. So the vertex is (2, −9).

2. Find the y-intercept

Set x = 0: f(0) = 0 − 0 − 5 = −5. The y-intercept is (0, −5).

3. Find the x-intercepts

Set f(x) = 0 and factor:

x2−4x−5=0⇒(x−5)(x+1)=0x^2 - 4x - 5 = 0 \quad\Rightarrow\quad (x - 5)(x + 1) = 0

So x = 5 or x = −1. The x-intercepts are (−1, 0) and (5, 0).

The parabola opens upward (a > 0) with vertex at (2, −9).

Downward-opening parabola: graph g(x) = −x² + 4x + 5.

Because a = −1 < 0, this parabola opens downward. It has a maximum point instead of a minimum. The vertex at (2, 9) is the highest the function ever reaches, so f(x) never goes above 9, making the range y ≤ 9. The orange curve shows the ∩ shape, with the gold vertex at the top. Compare this to the upward parabola above: the sign of “a” flips the entire shape.

4. Find the vertex

Here a = −1, b = 4, c = 5. The x-coordinate of the vertex is:

x=−b2a=−42(−1)=2x = -\frac{b}{2a} = -\frac{4}{2(-1)} = 2

Plug x = 2 back in: g(2) = −(2)² + 4(2) + 5 = −4 + 8 + 5 = 9. So the vertex is (2, 9), the highest point.

5. Find the y-intercept

Set x = 0: g(0) = 0 + 0 + 5 = 5. The y-intercept is (0, 5).

6. Find the x-intercepts

Set g(x) = 0: −x² + 4x + 5 = 0. Multiply both sides by −1 to get x² − 4x − 5 = 0, then factor:

(x−5)(x+1)=0(x - 5)(x + 1) = 0

So x = 5 or x = −1. The x-intercepts are (−1, 0) and (5, 0), same as the upward parabola, but the shape is flipped.

Real-World Application

Section titled “Real-World Application”

Quadratic functions model many real situations:

  • Projectile motion (height of a thrown ball over time)
  • Profit = revenue − cost (often quadratic)
  • Area of a rectangular garden with fixed perimeter
  • Bridge arches and satellite dishes (parabolic shapes)
  • Stopping distance of a car as speed increases

Example: a ball is thrown upward. Its height can be modeled by a quadratic function. The vertex represents the maximum height, a downward-opening parabola, just like the second example above.

Quiz

Section titled “Quiz”
For $f(x) = 2x^2 - 8x + 3$, does the parabola open upward or downward?
What is the x-coordinate of the vertex of $f(x) = x^2 + 6x - 7$?
The y-intercept of $f(x) = x^2 - 4x + 5$ is:
A quadratic function has a minimum point. This means:
Find the vertex of $f(x) = -x^2 + 4x + 1$.