# Parabolic Function — Definition, Formula, Graph, and Examples

A parabolic function is a degree-two function f(x)=ax²+bx+c (with a≠0) whose graph is a U-shaped curve called a parabola. This article defines the parabolic function, draws its graph with vertex and axis of symmetry marked, derives the vertex formula, gives its domain and range, and works through six examples — including how it differs from a plain quadratic equation.

## What Is a Parabolic Function?  
A **parabolic function** is a second-degree (quadratic) function of the form  
f(x)=ax²+bx+c, a≠0  
whose graph is a parabola — a symmetric, U-shaped curve. The condition a≠0 is essential: if a=0, the x² term vanishes and the function collapses into a straight line, not a parabola.

| Symbol  | Meaning  |
|---------|----------|
| a      | leading coefficient — controls width and opening direction (a≠0)  |
| b      | linear coefficient — shifts the axis of symmetry sideways  |
| c      | constant term — the y-intercept, f(0)=c  |
| vertex  | the turning point of the curve  |

Because two different inputs can give the same output (for instance, f(2) and f(−2) are equal when b=0), a parabolic function is a **many-to-one** function. This is the standard form used throughout [quadratic equations](/content/math/algebra/quadratic-equations/index.html) — the same ax²+bx+c written as a function rather than set equal to zero.

## What Does the Graph of a Parabolic Function Look Like?  
The graph is a U-shaped parabola. Three features describe it fully:

- **Opening direction.** If a>0, the parabola opens upward and the vertex is a minimum. If a<0, it opens downward and the vertex is a maximum.
- **Vertex.** The turning point, sitting at the very bottom (or top) of the curve.
- **Axis of symmetry.** A vertical line through the vertex that splits the parabola into two mirror halves.

The vertical line x=−b/2a is the [axis of symmetry](/content/math/algebra/axis-of-symmetry/index.html), and it always passes through the vertex.

## How Do You Find the Vertex of a Parabolic Function?  
The vertex sits on the axis of symmetry, so its x-coordinate is x=−b/2a. Substitute that back into the function to get the y-coordinate.  
Here is the one-line origin of that formula. The axis of symmetry sits exactly halfway between the two roots of ax²+bx+c=0. By the quadratic formula the roots are −b±√(b²−4ac)/2a, and their midpoint is −b/2a, because the ± part cancels. So the vertex's x-coordinate is −b/2a every time.

1. Compute x=−b/2a.
2. Substitute that x into f(x) to get the y-coordinate.
3. The vertex is (−b/2a, f(−b/2a)).

## Examples of Parabolic Function  
Six examples, from a clean vertex to a fraction-heavy case and a real-world projectile.

### Example 1
**Find the vertex of f(x)=x²−6x+5.**  
Here a=1, b=−6.

x=−b/2a=−(−6)/2(1)=3
f(3)=3²−6(3)+5=9−18+5=−4  
**Final answer:** vertex at (3,−4).

### Example 2
**A student finds the vertex of f(x)=2x²+8x+1 and writes x=−8/2=−4. Is that right?**  
_Wrong attempt._ The student uses x=−b/2, dividing only by 2 and forgetting the a.  
_Why it breaks._ That formula ignores the leading coefficient. With a=2, the denominator should be 2a=4, not 2.  
_Correct._ Use the full formula.

x=−b/2a=−8/2(2)=−8/4=−2
f(−2)=2(−2)²+8(−2)+1=8−16+1=−7  
**Final answer:** vertex at (−2,−7).

### Example 3
**Which way does f(x)=−3x²+12x−7 open, and what is its vertex?**  
Since a=−3<0, the parabola opens downward (vertex is a maximum).

x=−12/2(−3)=−12/−6=2
f(2)=−3(2)²+12(2)−7=−12+24−7=5  
**Final answer:** opens downward, vertex (maximum) at (2,5).

### Example 4
**State the domain and range of f(x)=x²+4.**  
A parabolic function accepts every real input, so the domain is all real numbers.  
The vertex is at (0,4) and the parabola opens upward, so outputs never drop below 4.  
**Final answer:** domain = ℝ; range = [4,∞).

### Example 5
**Find the vertex of f(x)=1/2x²−3x+4.**  
Here a=1/2, b=−3.

x=−b/2a=−(−3)/2(1/2)=−(−3)/1=3
f(3)=1/2(9)−3(3)+4=9/2−9+4=9/2−5=−1/2  
**Final answer:** vertex at (3,−1/2).

### Example 6
**A ball is thrown so its height (in metres) after t seconds is h(t)=−5t²+20t. When does it reach its highest point, and how high?**  
The path is a downward parabola (a=−5), so the peak is at the vertex.

t=−b/2a=−20/2(−5)=−20/−10=2  
h(2)=−5(2)²+20(2)=−20+40=20  
**Final answer:** the ball peaks at t=2 seconds, at a height of 20 metres.

## Why Parabolic Functions Matter  
The parabola earns its place across science because three unrelated-looking situations all produce the same curve.  
- **Projectile motion.** Under gravity alone, every thrown object traces a parabola. The vertex is the peak of the flight.  
- **Reflection and focusing.** A parabolic dish or mirror gathers parallel rays to a single focus.  
- **Optimization.** The vertex marks the maximum or minimum. Finding it is finding the best outcome.

The destination is optimization and the calculus of maxima and minima: the vertex you locate with −b/2a is the same extreme point a derivative will later find by setting the slope to zero. The [zeros of the function](/content/math/algebra/zeros-of-a-function/index.html) and its vertex together tell the whole story of the curve.

## Tripping Points to Avoid
### Mistake 1: Using x=−b/2 instead of x=−b/2a  
**Where it slips in:** Any parabolic function where a≠1.  
**Don't do this:** Drop the a from the denominator.  
**The correct way:** The axis of symmetry is x=−b/2a. The 2a is the whole denominator, and leaving off a silently shifts the vertex sideways.

### Mistake 2: Forgetting the opening direction depends on the sign of a  
**Where it slips in:** Stating whether the vertex is a maximum or a minimum.  
**Don't do this:** Assume every parabola opens upward.  
**The correct way:** Check the sign of a. If a>0 the parabola opens up (vertex is a minimum); if a<0 it opens down (vertex is a maximum).

### Mistake 3: Confusing a parabolic function with a quadratic equation  
**Where it slips in:** Switching between f(x)=ax²+bx+c and ax²+bx+c=0.  
**Don't do this:** Treat "finding the vertex" and "finding the roots" as the same task.  
**The correct way:** A parabolic function describes the whole curve (every x gives a y). A quadratic equation asks only where that curve equals zero (the roots).

## Practice Questions  
Try these, then check the answers below.  
1. Find the vertex of f(x)=x²+2x−8.
2. Decide which way f(x)=−x²+4x opens and where it peaks.
3. State the range of f(x)=3x²+6.
4. Find the axis of symmetry of f(x)=2x²−12x+5.
5. Give the y-intercept of f(x)=4x²−x+7.

### Answers  
**Answer to Question 1:** Vertex at (−1,−9).  
**Answer to Question 2:** Peaks at (2,4).  
**Answer to Question 3:** Range = [6,∞).  
**Answer to Question 4:** Axis of symmetry at x=3.  
**Answer to Question 5:** y-intercept at (0,7).

## Conclusion  
- A **parabolic function** is f(x)=ax²+bx+c with a≠0, and its graph is a U-shaped parabola.  
- The vertex sits at x=−b/2a; substitute back to get the y-coordinate.  
- The sign of a sets the opening direction: up (minimum) if a>0, down (maximum) if a<0.  
- The domain is all real numbers; the range starts (or ends) at the vertex's y-value.