Quadratic Functions and Equations
Explore the U-shaped curves and solve non-linear equations
You have spent a lot of time with linear equations - those straight-line relationships where the variable is never raised to a power higher than 1. Now it is time to level up. Quadratic functions introduce something new: the variable gets squared. That one small change - going from $x$ to $x^2$ - creates curves instead of lines, and those curves open up a whole new world of problems you can solve.
If you have ever thrown a ball and watched it arc through the air, you have seen a quadratic function in action. The path that ball follows is not a straight line - it curves up, reaches a peak, and curves back down. That U-shaped path is called a parabola, and understanding it gives you the tools to answer questions like “How high will it go?” and “Where will it land?”
Core Concepts
What Is a Quadratic Function?
A quadratic function is any function that can be written in the form:
$$f(x) = ax^2 + bx + c$$
where $a$, $b$, and $c$ are constants, and crucially, $a \neq 0$. That squared term is what makes it quadratic - “quad” comes from the Latin word for “square.”
The constants have names:
- $a$ is the leading coefficient (the number in front of $x^2$)
- $b$ is the linear coefficient (the number in front of $x$)
- $c$ is the constant term (the number with no variable)
For example, in $f(x) = 2x^2 - 5x + 3$:
- $a = 2$
- $b = -5$
- $c = 3$
The Shape: Parabolas
Every quadratic function graphs as a U-shaped curve called a parabola. Unlike the straight lines of linear functions, parabolas curve - they have a highest or lowest point, and they are symmetric.
The direction the parabola opens depends entirely on the leading coefficient $a$:
- If $a > 0$ (positive), the parabola opens upward like a smile
- If $a < 0$ (negative), the parabola opens downward like a frown
The size of $|a|$ affects how “wide” or “narrow” the parabola is:
- If $|a| > 1$, the parabola is narrower (steeper)
- If $|a| < 1$, the parabola is wider (flatter)
The Vertex: The Turning Point
The vertex is the highest or lowest point of the parabola - the point where the curve turns around. If the parabola opens upward, the vertex is the minimum point. If it opens downward, the vertex is the maximum point.
To find the x-coordinate of the vertex:
$$x = -\frac{b}{2a}$$
Once you have the x-coordinate, plug it back into the original function to find the y-coordinate.
This vertex formula is one of the most useful tools in working with quadratics. Memorize it.
The Axis of Symmetry
Parabolas are perfectly symmetric - if you folded the graph along a vertical line through the vertex, both halves would match exactly. This vertical line is called the axis of symmetry.
The equation of the axis of symmetry is:
$$x = -\frac{b}{2a}$$
Notice this is the same as the x-coordinate of the vertex. The axis of symmetry passes directly through the vertex, dividing the parabola into two mirror-image halves.
Zeros, Roots, and X-Intercepts
The zeros (also called roots or x-intercepts) of a quadratic function are the x-values where the function equals zero - the points where the parabola crosses the x-axis.
To find them, you set $f(x) = 0$ and solve:
$$ax^2 + bx + c = 0$$
A quadratic equation can have:
- Two distinct real solutions (parabola crosses x-axis twice)
- One repeated real solution (parabola touches x-axis at exactly one point - its vertex)
- No real solutions (parabola never crosses the x-axis)
There are three main methods for solving quadratic equations: factoring, square roots, and the quadratic formula.
Solving by Factoring: The Zero Product Property
When you can factor a quadratic, solving becomes straightforward thanks to the Zero Product Property:
If $ab = 0$, then $a = 0$ or $b = 0$ (or both).
In other words, if a product equals zero, at least one of the factors must be zero.
So if you can write $x^2 + 5x + 6 = 0$ as $(x + 2)(x + 3) = 0$, then either:
- $x + 2 = 0$, which gives $x = -2$, or
- $x + 3 = 0$, which gives $x = -3$
The solutions are $x = -2$ and $x = -3$.
This is why factoring is so powerful - it converts one hard equation into two easy ones.
Solving by Square Roots
Some quadratic equations have a simpler structure that allows you to solve by taking square roots directly. If your equation looks like $x^2 = k$ (where there is no linear term), you can solve by taking the square root of both sides:
$$x^2 = k$$ $$x = \pm\sqrt{k}$$
That $\pm$ symbol means “plus or minus” - you get two solutions, one positive and one negative.
For example, to solve $x^2 = 25$: $$x = \pm\sqrt{25} = \pm 5$$
So $x = 5$ or $x = -5$.
This also works for equations like $(x - 3)^2 = 16$: $$x - 3 = \pm 4$$ $$x = 3 + 4 = 7 \quad \text{or} \quad x = 3 - 4 = -1$$
The Quadratic Formula
When factoring is difficult or impossible, the quadratic formula always works. For any equation $ax^2 + bx + c = 0$:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
This formula gives you both solutions at once. The $\pm$ means you calculate it twice: once with the plus sign, once with the minus sign.
Yes, this formula looks intimidating at first. But it works every time, for every quadratic equation. Many students find it helpful to memorize it as a song or chant.
The Discriminant: Predicting the Solutions
The expression under the square root in the quadratic formula - the part $b^2 - 4ac$ - is called the discriminant. It tells you how many real solutions the equation has before you even finish solving:
- If $b^2 - 4ac > 0$: Two distinct real solutions (the square root gives a positive number)
- If $b^2 - 4ac = 0$: One repeated real solution (the square root is zero)
- If $b^2 - 4ac < 0$: No real solutions (you cannot take the square root of a negative)
The discriminant is a quick check that helps you know what to expect.
Graphing Quadratic Functions
To graph a quadratic function, gather these key features:
- Direction: Does it open up ($a > 0$) or down ($a < 0$)?
- Vertex: Calculate $x = -\frac{b}{2a}$, then find the corresponding y-value
- Axis of symmetry: The vertical line $x = -\frac{b}{2a}$
- Y-intercept: Set $x = 0$; the y-intercept is $c$
- X-intercepts (if they exist): Solve $ax^2 + bx + c = 0$
Plot the vertex, the y-intercept, and any x-intercepts. Use the axis of symmetry to plot additional symmetric points if needed. Then draw a smooth U-shaped curve through your points.
Notation and Terminology
| Term | Meaning | Example |
|---|---|---|
| Quadratic function | $f(x) = ax^2 + bx + c$ | $f(x) = x^2 - 4x + 3$ |
| Parabola | U-shaped graph of a quadratic | |
| Vertex | Highest or lowest point of the parabola | |
| Axis of symmetry | Vertical line through vertex | $x = -\frac{b}{2a}$ |
| Zeros/roots | X-intercepts; where $f(x) = 0$ | |
| Discriminant | $b^2 - 4ac$; determines number of solutions |
Examples
Solve $x^2 - 9 = 0$.
Solution:
First, isolate $x^2$: $$x^2 = 9$$
Take the square root of both sides (remembering both positive and negative roots): $$x = \pm\sqrt{9}$$ $$x = \pm 3$$
The solutions are $x = 3$ and $x = -3$.
Check: Does $3^2 - 9 = 0$? Yes, $9 - 9 = 0$ ✓
Does $(-3)^2 - 9 = 0$? Yes, $9 - 9 = 0$ ✓
Real-world connection: This is the structure you encounter when finding the side length of a square with a given area, or when working with the Pythagorean theorem.
Solve $(x - 3)(x + 5) = 0$.
Solution:
The equation is already factored. Apply the Zero Product Property:
Either $x - 3 = 0$ or $x + 5 = 0$
Solve each:
- $x - 3 = 0 \Rightarrow x = 3$
- $x + 5 = 0 \Rightarrow x = -5$
The solutions are $x = 3$ and $x = -5$.
Check: Does $(3 - 3)(3 + 5) = 0$? Yes, $0 \cdot 8 = 0$ ✓
Does $(-5 - 3)(-5 + 5) = 0$? Yes, $(-8) \cdot 0 = 0$ ✓
Why this works: Whenever you multiply any number by zero, the result is zero. So if a product equals zero, at least one factor must be zero.
Solve $x^2 + 2x - 15 = 0$ by factoring.
Solution:
We need two numbers that:
- Multiply to $-15$ (the constant term)
- Add to $2$ (the coefficient of $x$)
Think about factor pairs of 15: $(1, 15)$ and $(3, 5)$
Since the product is negative and the sum is positive, the larger number must be positive:
- $5$ and $-3$: $5 \times (-3) = -15$ ✓ and $5 + (-3) = 2$ ✓
Factor the equation: $$x^2 + 2x - 15 = (x + 5)(x - 3) = 0$$
Apply the Zero Product Property:
- $x + 5 = 0 \Rightarrow x = -5$
- $x - 3 = 0 \Rightarrow x = 3$
The solutions are $x = -5$ and $x = 3$.
Check: $(-5)^2 + 2(-5) - 15 = 25 - 10 - 15 = 0$ ✓
$3^2 + 2(3) - 15 = 9 + 6 - 15 = 0$ ✓
Find the vertex of $y = x^2 - 6x + 5$.
Solution:
For $y = x^2 - 6x + 5$, identify the coefficients:
- $a = 1$
- $b = -6$
- $c = 5$
Step 1: Find the x-coordinate of the vertex using the formula: $$x = -\frac{b}{2a} = -\frac{-6}{2(1)} = -\frac{-6}{2} = 3$$
Step 2: Find the y-coordinate by substituting $x = 3$ into the original equation: $$y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4$$
The vertex is at $(3, -4)$.
Interpretation: Since $a = 1 > 0$, this parabola opens upward. The vertex $(3, -4)$ is the lowest point on the graph - the minimum value of the function is $-4$, occurring when $x = 3$.
The axis of symmetry is the vertical line $x = 3$.
Solve $2x^2 + 5x - 3 = 0$ using the quadratic formula.
Solution:
Identify the coefficients:
- $a = 2$
- $b = 5$
- $c = -3$
Apply the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Step 1: Calculate the discriminant: $$b^2 - 4ac = (5)^2 - 4(2)(-3) = 25 - (-24) = 25 + 24 = 49$$
Step 2: Substitute into the formula: $$x = \frac{-5 \pm \sqrt{49}}{2(2)} = \frac{-5 \pm 7}{4}$$
Step 3: Calculate both solutions: $$x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}$$
$$x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3$$
The solutions are $x = \frac{1}{2}$ and $x = -3$.
Check: For $x = \frac{1}{2}$: $$2\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) - 3 = 2 \cdot \frac{1}{4} + \frac{5}{2} - 3 = \frac{1}{2} + \frac{5}{2} - 3 = 3 - 3 = 0$$ ✓
For $x = -3$: $$2(-3)^2 + 5(-3) - 3 = 2(9) - 15 - 3 = 18 - 18 = 0$$ ✓
Determine the number of real solutions for each equation without solving:
a) $x^2 - 4x + 4 = 0$
b) $x^2 + 2x + 5 = 0$
c) $3x^2 - 5x + 1 = 0$
Solution:
Calculate the discriminant $b^2 - 4ac$ for each:
a) For $x^2 - 4x + 4 = 0$ where $a = 1$, $b = -4$, $c = 4$: $$b^2 - 4ac = (-4)^2 - 4(1)(4) = 16 - 16 = 0$$
Since the discriminant equals zero, there is one repeated real solution.
(In fact, this is $(x - 2)^2 = 0$, so $x = 2$ is a double root.)
b) For $x^2 + 2x + 5 = 0$ where $a = 1$, $b = 2$, $c = 5$: $$b^2 - 4ac = (2)^2 - 4(1)(5) = 4 - 20 = -16$$
Since the discriminant is negative, there are no real solutions.
(The parabola never crosses the x-axis - it is entirely above it.)
c) For $3x^2 - 5x + 1 = 0$ where $a = 3$, $b = -5$, $c = 1$: $$b^2 - 4ac = (-5)^2 - 4(3)(1) = 25 - 12 = 13$$
Since the discriminant is positive, there are two distinct real solutions.
The discriminant summary:
- $b^2 - 4ac > 0$: Two real solutions
- $b^2 - 4ac = 0$: One real solution
- $b^2 - 4ac < 0$: No real solutions
Key Rules
The Quadratic Formula
For any equation $ax^2 + bx + c = 0$ where $a \neq 0$:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
This formula always works. When factoring is not obvious, go straight to this.
Vertex Formula
For $f(x) = ax^2 + bx + c$, the vertex is at:
$$\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$$
Find the x-coordinate with $-\frac{b}{2a}$, then substitute back to get the y-coordinate.
Zero Product Property
If $ab = 0$, then $a = 0$ or $b = 0$.
This is why factoring solves equations - once you have a product equal to zero, each factor can be set equal to zero and solved separately.
Discriminant Quick Reference
For $ax^2 + bx + c = 0$:
| Discriminant $b^2 - 4ac$ | Number of Real Solutions | Graph Behavior |
|---|---|---|
| Positive ($> 0$) | Two distinct solutions | Crosses x-axis twice |
| Zero ($= 0$) | One repeated solution | Touches x-axis once (at vertex) |
| Negative ($< 0$) | No real solutions | Never crosses x-axis |
Common Mistakes to Avoid
- Forgetting the $\pm$ when taking square roots - You always get two solutions (positive and negative)
- Sign errors in the quadratic formula - Note that it is $-b$, not $b$, at the start
- Dividing by $2a$ incorrectly - The entire numerator (both terms) gets divided by $2a$
- Stopping at the discriminant - The discriminant tells you how many solutions; you still need to finish the quadratic formula to find them
Real-World Applications
Projectile Motion
When you throw a ball upward, its height $h$ (in feet) after $t$ seconds follows a quadratic function:
$$h(t) = -16t^2 + v_0 t + h_0$$
where $v_0$ is the initial upward velocity (feet per second) and $h_0$ is the initial height. The negative coefficient on $t^2$ makes the parabola open downward - the ball goes up, reaches a peak, and comes back down.
The vertex gives you the maximum height and when it occurs. Setting $h(t) = 0$ and solving tells you when the ball hits the ground.
Maximum and Minimum Problems
Many optimization problems in business and science involve quadratics. If your profit function is $P(x) = -2x^2 + 100x - 800$ (where $x$ is the number of items produced), the vertex tells you exactly how many items to produce for maximum profit.
Since $a = -2 < 0$, the parabola opens downward, meaning the vertex is a maximum point. Use $x = -\frac{b}{2a}$ to find the optimal production level.
Architecture
Parabolic arches appear throughout architecture - from ancient Roman aqueducts to modern bridges. The parabola is structurally strong because it distributes weight evenly. Architects and engineers use quadratic equations to design these curves with precise dimensions.
Stopping Distance
The distance required for a car to stop depends on speed in a quadratic way. Doubling your speed does not double your stopping distance - it roughly quadruples it. This is because kinetic energy (and thus the work needed to stop) depends on velocity squared:
$$d = \frac{v^2}{2\mu g}$$
Understanding this quadratic relationship explains why speeding is so dangerous.
Self-Test Problems
Problem 1: Solve $x^2 = 49$.
Show Answer
Take the square root of both sides: $$x = \pm\sqrt{49} = \pm 7$$
The solutions are $x = 7$ and $x = -7$.
Check: $(7)^2 = 49$ ✓ and $(-7)^2 = 49$ ✓
Problem 2: Solve $(x + 4)(x - 2) = 0$.
Show Answer
By the Zero Product Property:
- $x + 4 = 0 \Rightarrow x = -4$
- $x - 2 = 0 \Rightarrow x = 2$
The solutions are $x = -4$ and $x = 2$.
Check: $(-4 + 4)(-4 - 2) = 0 \cdot (-6) = 0$ ✓ and $(2 + 4)(2 - 2) = 6 \cdot 0 = 0$ ✓
Problem 3: Solve $x^2 - x - 12 = 0$ by factoring.
Show Answer
Find two numbers that multiply to $-12$ and add to $-1$:
- $3$ and $-4$: $3 \times (-4) = -12$ and $3 + (-4) = -1$ ✓
Factor: $(x + 3)(x - 4) = 0$
Solutions: $x = -3$ or $x = 4$
Check: $(-3)^2 - (-3) - 12 = 9 + 3 - 12 = 0$ ✓
$(4)^2 - 4 - 12 = 16 - 4 - 12 = 0$ ✓
Problem 4: Find the vertex and axis of symmetry for $y = x^2 + 4x - 5$.
Show Answer
For $y = x^2 + 4x - 5$: $a = 1$, $b = 4$, $c = -5$
Axis of symmetry: $x = -\frac{b}{2a} = -\frac{4}{2(1)} = -2$
Vertex x-coordinate: $x = -2$
Vertex y-coordinate: $y = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9$
The vertex is $(-2, -9)$ and the axis of symmetry is $x = -2$.
Since $a > 0$, the parabola opens upward and $(-2, -9)$ is the minimum point.
Problem 5: Solve $x^2 + 3x - 2 = 0$ using the quadratic formula.
Show Answer
For $x^2 + 3x - 2 = 0$: $a = 1$, $b = 3$, $c = -2$
Discriminant: $b^2 - 4ac = 9 - 4(1)(-2) = 9 + 8 = 17$
Quadratic formula: $$x = \frac{-3 \pm \sqrt{17}}{2(1)} = \frac{-3 \pm \sqrt{17}}{2}$$
The solutions are $x = \frac{-3 + \sqrt{17}}{2} \approx 0.56$ and $x = \frac{-3 - \sqrt{17}}{2} \approx -3.56$
Problem 6: Without solving, determine how many real solutions $2x^2 + 4x + 5 = 0$ has.
Show Answer
Calculate the discriminant: $$b^2 - 4ac = (4)^2 - 4(2)(5) = 16 - 40 = -24$$
Since the discriminant is negative ($-24 < 0$), the equation has no real solutions.
The parabola $y = 2x^2 + 4x + 5$ opens upward (since $a = 2 > 0$) and its vertex is above the x-axis, so it never crosses the x-axis.
Summary
- A quadratic function has the form $f(x) = ax^2 + bx + c$ where $a \neq 0$. The squared term is what makes it quadratic.
- The graph of every quadratic is a parabola - a U-shaped curve that opens upward if $a > 0$ and downward if $a < 0$.
- The vertex is the turning point of the parabola. Find its x-coordinate with $x = -\frac{b}{2a}$, then substitute back to get the y-coordinate.
- The axis of symmetry is the vertical line $x = -\frac{b}{2a}$ that passes through the vertex.
- To solve quadratic equations, you have three main tools:
- Factoring with the Zero Product Property (when factoring is possible)
- Square roots (when there is no linear term)
- Quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ (works for every quadratic)
- The discriminant $b^2 - 4ac$ tells you the number of real solutions: positive means two, zero means one, negative means none.
- Quadratic relationships appear everywhere: projectile motion, optimization problems, architecture, and stopping distance are just a few examples.
You have now graduated from straight lines to curves. The techniques you have learned here - especially the quadratic formula and vertex formula - will serve you throughout algebra, precalculus, and beyond. When you encounter parabolas in future courses, you will already understand how they work.