This page covers quadratic functions from the formulas up. The first section lists the key formulas and properties: vertex, axis of symmetry, maximum/minimum, and range. The second section then provides 15 fully solved practice problems including interval analysis and reverse-engineering a quadratic from given conditions. Ideal for algebra and precalculus students.
A quadratic function has the form \[ f(x) = ax^2 + bx + c, \qquad a \neq 0. \] Its graph is a parabola. The table below summarizes the standard formulas and properties used throughout the problems.
The vertex is the highest or lowest point on the parabola. Its \( x \)-coordinate is \[ h = -\frac{b}{2a}, \] and its \( y \)-coordinate is \( k = f(h) \). The axis of symmetry is the vertical line \[ x = -\frac{b}{2a}. \]
| Sign of \(a\) | Direction | Vertex is a… | Range |
|---|---|---|---|
| \(a > 0\) | opens upward | minimum | \( [k, +\infty) \) |
| \(a < 0\) | opens downward | maximum | \( (-\infty, k] \) |
Here \( h = -\tfrac{b}{2a} \).
When the vertex \( (h, k) \) is known, start from vertex form \( f(x) = a(x - h)^2 + k \) and use one additional point to solve for \( a \). When the intervals of increase/decrease are given, the turning point is the axis of symmetry.
Find the maximum or minimum value of \( f(x) = 2x^2 + 3x - 5 \).
\(a = 2 > 0\), so the parabola opens upward → minimum at the vertex.
\[ x = -\frac{3}{2(2)} = -\frac{3}{4} \] \[ f\!\left(-\frac{3}{4}\right) = 2\!\left(\frac{9}{16}\right) - \frac{9}{4} - 5 = \frac{9}{8} - \frac{18}{8} - \frac{40}{8} = -\frac{49}{8} \]Answer: Minimum value \( -\dfrac{49}{8} \)
Find the range of \( f(x) = -x^2 + 4x - 5 \).
\(a = -1 < 0\), opens downward → maximum at the vertex.
\[ x = -\frac{4}{2(-1)} = 2, \qquad f(2) = -4 + 8 - 5 = -1 \]Answer: Range is \( (-\infty, -1] \)
Find the vertex of the graph of \( f(x) = 3x^2 + 6x - 10 \).
Answer: Vertex \( (-1, -13) \)
Find the intervals of increase and decrease of \( f(x) = 6x^2 + x - 2 \).
\(a = 6 > 0\), opens upward → decrease before vertex, increase after.
\[ x = -\frac{1}{2(6)} = -\frac{1}{12} \]Answer: Decreases on \( (-\infty, -\tfrac{1}{12}) \), increases on \( (-\tfrac{1}{12}, +\infty) \)
Find the axis of symmetry of \( f(x) = -2x^2 - x - 2 \).
Answer: \( x = -\dfrac{1}{4} \)
Find the maximum or minimum value of \( f(x) = -3x^2 + 9x \).
\(a = -3 < 0\), so maximum at the vertex.
\[ x = -\frac{9}{2(-3)} = \frac{3}{2} \] \[ f\!\left(\frac{3}{2}\right) = -3 \cdot \frac{9}{4} + 9 \cdot \frac{3}{2} = -\frac{27}{4} + \frac{54}{4} = \frac{27}{4} \]Answer: Maximum value \( \dfrac{27}{4} \)
Find the range of \( f(x) = x^2 + 5x - 2 \).
\(a = 1 > 0\), opens upward → minimum at vertex.
\[ x = -\frac{5}{2}, \qquad f\!\left(-\frac{5}{2}\right) = \frac{25}{4} - \frac{25}{2} - 2 = -\frac{33}{4} \]Answer: Range \( \left[-\dfrac{33}{4}, +\infty\right) \)
Find the vertex of the graph of \( f(x) = -x^2 \).
Standard parent reflected over the \(x\)-axis. The vertex is at the origin.
Answer: Vertex \( (0, 0) \)
Find the intervals of increase and decrease of \( f(x) = -\tfrac{1}{2}x^2 + 1.1x - 2.3 \).
\(a = -0.5 < 0\), opens downward → increase before vertex, decrease after.
\[ x = -\frac{1.1}{2(-0.5)} = 1.1 \]Answer: Increases on \( (-\infty, 1.1) \), decreases on \( (1.1, +\infty) \)
Find the axis of symmetry of \( f(x) = -\tfrac{1}{2}x^2 + 1.1x - 2.3 \).
From the previous question, \(x = 1.1\).
Answer: \( x = 1.1 \)
Find the range of \( f(x) = -(x - 2)^2 + 2x + 4 \).
Expand into standard form:
\[ f(x) = -(x^2 - 4x + 4) + 2x + 4 = -x^2 + 4x - 4 + 2x + 4 = -x^2 + 6x \]\(a = -1 < 0\), so maximum at vertex:
\[ x = -\frac{6}{2(-1)} = 3, \qquad f(3) = -9 + 18 = 9 \]Answer: Range \( (-\infty, 9] \)
Find the vertex of \( f(x) = -(x + 4)^2 + 4x - 2 \).
Expand into standard form:
\[ f(x) = -(x^2 + 8x + 16) + 4x - 2 = -x^2 - 8x - 16 + 4x - 2 = -x^2 - 4x - 18 \] \[ x = -\frac{-4}{2(-1)} = -2, \qquad f(-2) = -4 + 8 - 18 = -14 \]Answer: Vertex \( (-2, -14) \)
Find the quadratic function \(f\) whose maximum value is \(-3\), axis of symmetry is \(x = 2\), and \(f(0) = -9\).
Vertex \( (h, k) = (2, -3) \), so start from vertex form:
\[ f(x) = a(x - 2)^2 - 3 \]Use the point \( (0, -9) \):
\[ -9 = a(0 - 2)^2 - 3 = 4a - 3 \implies -6 = 4a \implies a = -\frac{3}{2} \]Answer: \( f(x) = -\dfrac{3}{2}(x - 2)^2 - 3 \)
Find the quadratic function \(f\) that increases on \((-\infty, -2)\), decreases on \((-2, +\infty)\), and satisfies \(f(0) = 23\) and \(f(1) = 8\).
The turning point is at \(x = -2\), so the axis of symmetry is \(x = -2\) and the vertex form is \(f(x) = a(x + 2)^2 + k\).
Using the two points:
\[ f(0) = 23 \implies 4a + k = 23 \] \[ f(1) = 8 \implies 9a + k = 8 \]Subtract the first from the second: \(5a = -15 \Rightarrow a = -3\). Then \(4(-3) + k = 23 \Rightarrow k = 35\).
Answer: \( f(x) = -3(x + 2)^2 + 35 \)
Find the quadratic function \(f\) whose minimum value is \(2\), axis of symmetry is \(x = -3\), and \(f(2) = 1\).
Vertex \((-3, 2)\) opening upward means \(f(x) \ge 2\) for all \(x\).
But \(f(2) = 1 < 2\), a contradiction. No such function exists.
Answer: No such quadratic function exists.
\(f(x) = ax^2 + bx + c\), with \(a \neq 0\). The graph is a parabola with vertex \(x\)-coordinate \(-\tfrac{b}{2a}\).
Compute \(h = -\tfrac{b}{2a}\), then \(k = f(h)\). The vertex is \((h, k)\).
If \(a > 0\), the parabola opens upward and has a minimum. If \(a < 0\), it opens downward and has a maximum.
If \(a > 0\), the range is \([k, +\infty)\). If \(a < 0\), it's \((-\infty, k]\), where \(k\) is the \(y\)-coordinate of the vertex.
Use vertex form \(f(x) = a(x - h)^2 + k\), substitute the vertex \((h,k)\) and the given point, then solve for \(a\).