Quadratic Functions in Vertex Form
Quadratic functions in vertex form are written as \( f(x) = a(x - h)^2 + k \). This form makes the vertex \((h, k)\) visible directly, so the graph's key properties — vertex, axis of symmetry, maximum/minimum, and intercepts — are easy to read off.
The graph of a quadratic function is a parabola, a U-shaped curve. Changing \(a\), \(h\), and \(k\) transforms the parabola. You can see this interactively in the applet below.
A - Vertex, Maximum, and Minimum Values
For \( f(x) = a(x - h)^2 + k \), the squared term \((x - h)^2\) is always non-negative: \[ (x - h)^2 \ge 0. \] Multiplying by \(a\) gives two cases:
- Case 1: \(a > 0\) (positive)
\[ a(x - h)^2 \ge 0 \;\Rightarrow\; a(x - h)^2 + k \ge k \;\Rightarrow\; f(x) \ge k. \] So \(k\) is the minimum value of \(f\), attained at the vertex \((h, k)\). - Case 2: \(a < 0\) (negative)
\[ a(x - h)^2 \le 0 \;\Rightarrow\; a(x - h)^2 + k \le k \;\Rightarrow\; f(x) \le k. \] So \(k\) is the maximum value of \(f\), again at the vertex \((h, k)\).
Thus, the vertex \((h, k)\) is a minimum point when \(a > 0\) and a maximum point when \(a < 0\). The vertical line \(x = h\) is the axis of symmetry.
Example: Identify the Vertex and Extremum Type
Find the vertex of each function and classify it as a minimum or maximum.
- \( f(x) = -(x + 2)^2 - 1 \)
- \( f(x) = -x^2 + 2 \)
- \( f(x) = 2(x - 3)^2 \)
Show solution
- Rewrite as \( f(x) = -(x - (-2))^2 - 1 \). Here \(a = -1\), \(h = -2\), \(k = -1\). Vertex: \((-2, -1)\), a maximum.
- Rewrite as \( f(x) = -(x - 0)^2 + 2 \). Here \(a = -1\), \(h = 0\), \(k = 2\). Vertex: \((0, 2)\), a maximum.
- Rewrite as \( f(x) = 2(x - 3)^2 + 0 \). Here \(a = 2\), \(h = 3\), \(k = 0\). Vertex: \((3, 0)\), a minimum.
B - X-Intercepts of a Quadratic Function in Vertex Form
The x-intercepts are the real solutions (if any) of: \[ a(x - h)^2 + k = 0. \] Solving step by step: \[ \begin{aligned} a(x - h)^2 &= -k \\ (x - h)^2 &= -\frac{k}{a} \\ x - h &= \pm \sqrt{-\frac{k}{a}} \\ x &= h \pm \sqrt{-\frac{k}{a}}. \end{aligned} \] Real solutions exist only when \(-\frac{k}{a} \ge 0\).
Example: Finding X-Intercepts
Find the x-intercepts for each function.
- \( f(x) = -2(x - 3)^2 + 2 \)
- \( g(x) = -(x + 2)^2 \)
- \( h(x) = 4(x - 1)^2 + 5 \)
Show solution
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Solve \(-2(x - 3)^2 + 2 = 0\):
\[ \begin{aligned} -2(x - 3)^2 &= -2 \\ (x - 3)^2 &= 1 \\ x - 3 &= \pm 1 \\ x &= 4 \quad \text{or} \quad x = 2. \end{aligned} \] X-intercepts: \((4, 0)\) and \((2, 0)\). -
Solve \(-(x + 2)^2 = 0\):
\[ x + 2 = 0 \;\Rightarrow\; x = -2. \] One x-intercept at \((-2, 0)\). -
Solve \(4(x - 1)^2 + 5 = 0\):
\[ 4(x - 1)^2 = -5 \;\Rightarrow\; (x - 1)^2 = -\frac{5}{4}. \] No real solution. Hence, no x-intercepts.
C - Y-Intercept of a Quadratic Function in Vertex Form
The y-intercept is the value of \( f(0) \):
\[ f(0) = a(0 - h)^2 + k = ah^2 + k. \]So the parabola always crosses the y-axis at \( (0,\; ah^2 + k) \). This matches \( c = ah^2 + k \) when the function is written in general form \( f(x) = ax^2 + bx + c \).
Example: Finding the Y-Intercept
For \( f(x) = -2(x - 3)^2 + 2 \), the y-intercept is \( f(0) = -2(0 - 3)^2 + 2 = -18 + 2 = -16 \), so the intercept is \( (0, -16) \).
D - Interactive Applet: Explore the Vertex Form
Use the sliders (or type values) to change \(a\), \(h\), and \(k\), then click Draw. Watch how the vertex, the x-intercepts, and the y-intercept change.
E - Converting Vertex Form to General Form
To convert \( f(x) = a(x - h)^2 + k \) to general form \( f(x) = ax^2 + bx + c \), expand the square and combine like terms: \[ \begin{aligned} f(x) &= a(x - h)^2 + k \\ &= a(x^2 - 2hx + h^2) + k \\ &= ax^2 - 2ah\,x + (ah^2 + k). \end{aligned} \] Therefore \( b = -2ah \) and \( c = ah^2 + k \).
Example: Conversion
Rewrite \( f(x) = -(x - 2)^2 - 4 \) into general form.
Show solution
So \( a = -1 \), \( b = 4 \), \( c = -8 \).