In short

The quadratic function f(x) = a(x - h)^2 + k takes every value from k to +\infty when a > 0 (range [k, \infty)) and every value from -\infty to k when a < 0 (range (-\infty, k]). Over a closed interval [p, q], the range is bounded by the function values at the endpoints and (if it falls inside the interval) the vertex. The sign of f(x) — where it is positive, where it is negative — is controlled by the roots and the sign of a.

Take f(x) = x^2. Feed in any real number and the output is non-negative. You get 0 at x = 0, 1 at x = \pm 1, 4 at x = \pm 2, 100 at x = \pm 10. The output can be 0 or any positive number — but it can never be -1, or -7, or any negative number. The number -3 is simply not in the output of this function.

The set of all possible outputs of a function is called its range. For f(x) = x^2, the range is [0, \infty) — all non-negative reals. The range tells you what values the function can actually produce, no matter what input you give it.

For a quadratic function, the range is controlled entirely by two things: the sign of a (which decides whether the parabola opens up or down) and the vertex value k (which is the extreme output). Finding the range is one of the most common questions in algebra and competitive exams, and the answer almost always comes from the vertex.

Range over all reals

From the vertex form, every quadratic can be written as f(x) = a(x - h)^2 + k, where (h, k) is the vertex. The squared term (x - h)^2 is always \ge 0, so:

Range of a quadratic function

For f(x) = ax^2 + bx + c with vertex value k = c - \dfrac{b^2}{4a}:

\text{Range} = \begin{cases} [k,\, +\infty) & \text{if } a > 0 \\[6pt] (-\infty,\, k] & \text{if } a < 0 \end{cases}

The value k = -\dfrac{D}{4a}, where D = b^2 - 4ac is the discriminant.

The formula k = -D/(4a) connects the range to the discriminant. When a > 0 and D > 0 (two real roots), k < 0 — the vertex is below the axis, the minimum is negative, and the function takes some negative values. When a > 0 and D < 0 (no real roots), k > 0 — the entire parabola floats above the axis, and every output is positive.

Range of an upward-opening parabolaA parabola y equals x squared minus 4x plus 5 opening upward with vertex at 2 comma 1. A shaded vertical strip on the right side of the graph represents the range, extending from y equals 1 upward to infinity. A horizontal dashed line at y equals 1 marks the minimum value. The range bracket is labelled range equals 1 to infinity. x y 1 2 3 4 1 3 5 [1, ∞) vertex (2, 1) min = 1
The parabola $y = x^2 - 4x + 5 = (x - 2)^2 + 1$ opens upward with vertex at $(2, 1)$. The function value is always at least $1$ — it can never dip below the vertex. The range, shown as the shaded strip on the right, is $[1, \infty)$.

Range over a closed interval

The story changes when you restrict the input to a closed interval [p, q]. Now the function doesn't have infinite arms — it starts at x = p and ends at x = q. The range on this interval is the set of all outputs between the smallest and largest values of f on [p, q].

A continuous function on a closed interval always achieves its maximum and minimum (this is the extreme value theorem, which you will meet properly in calculus). For a quadratic, the extreme values can only occur at three places:

  1. The left endpoint x = p
  2. The right endpoint x = q
  3. The vertex x = h (but only if h \in [p, q])

So the procedure is:

Step 1. Find h = -b/(2a).

Step 2. Check whether h lies in [p, q].

Step 3. Compute f(p), f(q), and (if h \in [p, q]) f(h) = k.

Step 4. The range on [p, q] is [\text{smallest of these}, \text{largest of these}].

When the vertex falls outside the interval, the function is monotonic on [p, q] — it is either entirely increasing or entirely decreasing. In that case, the range is simply the interval between f(p) and f(q).

When the vertex falls inside [p, q], the function decreases from f(p) to f(h) and then increases from f(h) to f(q) (for a > 0), so the minimum is at the vertex and the maximum is at whichever endpoint gives the larger value.

Range of a quadratic over two different intervalsTwo panels side by side. The left panel shows the parabola y equals x squared minus 4x plus 5 on the interval from 0 to 5. The vertex at x equals 2 is inside the interval, so the minimum is 1 at the vertex and the maximum is 10 at x equals 5. The right panel shows the same parabola on the interval from 3 to 5. The vertex at x equals 2 is outside this interval, so the function is increasing throughout, and the range is from f of 3 equals 2 to f of 5 equals 10. interval [0, 5] 1 2 3 4 5 min = 1 max = 10 range: [1, 10] interval [3, 5] 1 2 3 4 5 min = 2 max = 10 range: [2, 10]
Same parabola $y = (x - 2)^2 + 1$, two different intervals. **Left:** on $[0, 5]$, the vertex $x = 2$ is inside the interval, so the minimum is at the vertex ($f(2) = 1$) and the maximum is at the farther endpoint ($f(5) = 10$). **Right:** on $[3, 5]$, the vertex is outside the interval, so the function is increasing throughout and the range is simply $[f(3), f(5)] = [2, 10]$.

Sign of a quadratic expression

A closely related question: for which values of x is f(x) > 0, and for which is f(x) < 0? This is the sign analysis of the quadratic, and the answer comes directly from the roots and the leading coefficient.

Case 1: D > 0 (two distinct real roots \alpha < \beta). The parabola crosses the axis at \alpha and \beta.

The sign between the roots is opposite to the sign outside the roots.

Case 2: D = 0 (one repeated root \alpha). The parabola touches the axis at \alpha without crossing.

The function has a definite sign everywhere except at the single root.

Case 3: D < 0 (no real roots). The parabola never touches the axis.

Sign of a quadratic in three cases based on the discriminantThree parabolas stacked vertically. The top one has two roots and the region between the roots is shaded to show where the function is negative. The middle one touches the axis at one point and is entirely non-negative. The bottom one floats above the axis and is entirely positive. Sign of f(x) = ax² + bx + c when a > 0 α β f > 0 f < 0 f > 0 D > 0 α f > 0 f > 0 D = 0 f > 0 everywhere D < 0
Sign chart for $f(x) = ax^2 + bx + c$ when $a > 0$. **Top:** $D > 0$, the function is negative between the roots (shaded) and positive outside. **Middle:** $D = 0$, the function is non-negative everywhere, touching zero only at the repeated root. **Bottom:** $D < 0$, the function is strictly positive for every $x$ — the parabola never reaches the axis.

A useful shortcut: when a > 0 and D < 0, the expression ax^2 + bx + c is positive definite — it is positive for every real x. When a < 0 and D < 0, it is negative definite — negative for every real x. These terms come up frequently in JEE problems.

Graphical analysis

The graph is the most reliable tool for range and sign questions. Here is a summary of what to read from the graph of f(x) = a(x - h)^2 + k:

Feature How to read it
Opens up or down Sign of a: up if a > 0, down if a < 0
Vertex (h, k) from vertex form
Axis of symmetry x = h
Minimum or maximum value k (minimum if a > 0, maximum if a < 0)
Range over \mathbb{R} [k, \infty) if a > 0; (-\infty, k] if a < 0
Roots Where the curve crosses the axis (if it does)
Sign of f(x) Above the axis means positive; below means negative

All of these are visible in a single well-drawn graph.

Interactive: exploring range over intervals

Drag the two red endpoints below to change the interval [p, q]. The readout shows the minimum and maximum values of f(x) = (x - 3)^2 - 2 on your chosen interval. When the vertex x = 3 is between the endpoints, the minimum drops to -2. When both endpoints are on the same side of the vertex, the minimum is at the nearer endpoint.

Interactive range explorer for a quadratic on a variable intervalAn interactive graph of the parabola y equals the quantity x minus 3 squared minus 2. Two draggable red points sit on the curve. A readout panel shows the function values at both endpoints and the minimum and maximum over the interval between them. drag the endpoints to change the interval
Drag either red point to change the interval. Watch how $f(p)$ and $f(q)$ change. When the vertex at $x = 3$ lies between the two endpoints, the minimum value on the interval is $-2$ (at the vertex). When both points are to the right of the vertex, the minimum is at the left endpoint and both $f(p)$ and $f(q)$ are above $-2$.

Example 1: Find the range of $f(x) = -x^2 + 6x - 5$ over all real numbers

Step 1. Identify a = -1, b = 6, c = -5. Since a < 0, the parabola opens downward — the function has a maximum, no minimum.

Why: a negative leading coefficient means the arms go to -\infty. The vertex is the peak.

Step 2. Find the vertex. h = -b/(2a) = -6/(2 \cdot (-1)) = 3.

k = f(3) = -9 + 18 - 5 = 4

Why: the vertex is at (3, 4). Since a < 0, this is the maximum point.

Step 3. Write vertex form to confirm: f(x) = -(x^2 - 6x) - 5 = -(x^2 - 6x + 9 - 9) - 5 = -(x - 3)^2 + 9 - 5 = -(x - 3)^2 + 4.

Why: vertex form -(x - 3)^2 + 4 makes it clear that -(x - 3)^2 \le 0, so f(x) \le 4 for every x.

Step 4. State the range.

Since f(x) \le 4 for all x, and f(3) = 4 (so the value 4 is achieved), and f(x) \to -\infty as x \to \pm\infty, the range is (-\infty, 4].

Result. Range of f(x) = -x^2 + 6x - 5 is (-\infty, 4].

Downward parabola with vertex at 3 comma 4 showing the range is negative infinity to 4The parabola y equals negative x squared plus 6x minus 5 opens downward. The vertex at 3 comma 4 is marked with a filled red circle. A horizontal dashed line at y equals 4 shows the maximum value. The roots are at x equals 1 and x equals 5. x y 1 2 3 4 5 vertex (3, 4) max = 4 range: (−∞, 4]
The parabola $y = -(x - 3)^2 + 4$ opens downward. The vertex at $(3, 4)$ is the highest point — the function can never exceed $4$. The range is every value from $-\infty$ up to and including $4$.

The vertex form -(x - 3)^2 + 4 settles the range in one glance: the squared term can only subtract from 4, so 4 is the ceiling.

Example 2: Find the range of $f(x) = x^2 - 4x + 5$ on the interval $[0, 5]$

Step 1. Find the vertex. h = -(-4)/(2 \cdot 1) = 2. k = f(2) = 4 - 8 + 5 = 1. Vertex: (2, 1).

Why: the vertex at x = 2 lies inside the interval [0, 5], so it will contribute to the range.

Step 2. Compute f at the endpoints. f(0) = 0 - 0 + 5 = 5. f(5) = 25 - 20 + 5 = 10.

Why: on a closed interval, the extreme values occur either at the vertex or at an endpoint. All three candidates must be checked.

Step 3. Compare the three values: f(0) = 5, f(2) = 1, f(5) = 10.

The smallest is f(2) = 1 (at the vertex). The largest is f(5) = 10 (at the right endpoint).

Why: the vertex gives the minimum because a > 0 and the vertex is inside the interval. The farther endpoint x = 5 is 3 units from the vertex while x = 0 is only 2 units away, so the function climbs higher on the right side.

Step 4. State the range.

Result. Range of f(x) = x^2 - 4x + 5 on [0, 5] is [1, 10].

Parabola on the interval 0 to 5 with minimum at vertex and maximum at right endpointThe parabola y equals x squared minus 4x plus 5 is shown on the interval from 0 to 5. The vertex at 2 comma 1 is the lowest point, marked with a filled circle. The endpoint values f of 0 equals 5 and f of 5 equals 10 are marked. The range bracket on the right shows values from 1 to 10. x y 1 2 3 4 5 1 5 9 min: f(2) = 1 max: f(5) = 10 f(0) = 5
On $[0, 5]$, the parabola $y = x^2 - 4x + 5$ dips to its vertex minimum $1$ at $x = 2$, then rises to $10$ at the right endpoint $x = 5$. The left endpoint value $f(0) = 5$ is neither the minimum nor the maximum. The range is $[1, 10]$.

Notice how the range on a restricted interval is a closed interval [1, 10] — both endpoints are achieved. Over all reals, the range would be [1, \infty), but restricting the domain also restricts the range. The right endpoint x = 5 is farther from the vertex than the left endpoint x = 0, so it produces the larger function value.

Common confusions

Going deeper

If you came here to learn how to find the range of a quadratic and analyse its sign, you have it — you can stop here. The rest connects the range to the idea of invertibility and to a technique for bounding expressions.

Why range matters for invertibility

A function is invertible only on a domain where it is one-to-one — each output comes from exactly one input. A quadratic f(x) = a(x - h)^2 + k is not one-to-one on all of \mathbb{R} (because f(h + d) = f(h - d) for every d, so each output other than k comes from two inputs). But if you restrict the domain to [h, \infty) (just the right half of the parabola), the function becomes one-to-one. Its range on that restricted domain is [k, \infty) (when a > 0), and the inverse function is

f^{-1}(y) = h + \sqrt{\frac{y - k}{a}}

defined for y \ge k. The range of f becomes the domain of f^{-1}.

Bounding quadratic expressions

A common competitive-exam technique: given an expression like \frac{x^2 + 2x + 3}{x^2 + 2x + 5}, find its range over all real x. Set the expression equal to y and rearrange into a quadratic in x:

x^2 + 2x + 3 = y(x^2 + 2x + 5)
(1 - y)x^2 + (2 - 2y)x + (3 - 5y) = 0

For real x to exist, the discriminant of this quadratic (in x) must be \ge 0:

D_x = (2 - 2y)^2 - 4(1 - y)(3 - 5y) \ge 0

Expanding and simplifying gives a quadratic inequality in y, and the solution set of that inequality is exactly the range of the original expression. This technique — "let the expression equal y, demand D \ge 0 in x" — is one of the most powerful range-finding tools in algebra.

The range and the discriminant, in reverse

The connection works both ways. You can use the range to prove things about the discriminant. If f(x) = ax^2 + bx + c with a > 0 has minimum value k = -D/(4a) > 0, then D < 0 (since a > 0 and k > 0 force -D/(4a) > 0, hence D < 0). This means f(x) > 0 for all x — the function has no real roots. So "positive range" and "negative discriminant" are two ways of saying the same thing. The range and the discriminant are not independent facts about a quadratic — each determines the other.

Where this leads next