Quadratic Functions in Vertex Form

Quadratic functions in vertex form are expressed as: \[ f(x) = a(x - h)^2 + k \] This form highlights the vertex \((h, k)\) of the parabola, making it easy to analyze the graph's properties such as the vertex, axis of symmetry, and intercepts.

The graph of a quadratic function is a parabola, a U-shaped curve. By changing the coefficients \(a\), \(h\), and \(k\), you can explore how the parabola transforms.

A - Vertex, Maximum, and Minimum Values

For \( f(x) = a(x - h)^2 + k \), the term \((x - h)^2\) is always non-negative: \[ (x - h)^2 \ge 0 \] Multiplying by coefficient \(a\) leads to two cases:

  1. Case 1: \(a > 0\) (positive)
    \[ a(x - h)^2 \ge 0 \Rightarrow a(x - h)^2 + k \ge k \Rightarrow f(x) \ge k \] Hence, \(k\) is the minimum value of \(f\), occurring at the vertex \((h, k)\).
  2. Case 2: \(a < 0\) (negative)
    \[ a(x - h)^2 \le 0 \Rightarrow a(x - h)^2 + k \le k \Rightarrow f(x) \le k \] Hence, \(k\) is the maximum value of \(f\), also at the vertex \((h, k)\).

Thus, the vertex \((h, k)\) is a minimum point if \(a > 0\) and a maximum point if \(a < 0\).

Example: Identify the Vertex and Extremum Type

Find the vertex of each function and classify it as a minimum or maximum point.

  1. \( f(x) = -(x + 2)^2 - 1 \)
  2. \( f(x) = -x^2 + 2 \)
  3. \( f(x) = 2(x - 3)^2 \)

Solution:

  1. Rewrite as \( f(x) = -(x - (-2))^2 - 1 \). Here \(a = -1\), \(h = -2\), \(k = -1\). Vertex: \((-2, -1)\), a maximum (since \(a < 0\)).
  2. Rewrite as \( f(x) = -(x - 0)^2 + 2 \). Here \(a = -1\), \(h = 0\), \(k = 2\). Vertex: \((0, 2)\), a maximum.
  3. Rewrite as \( f(x) = 2(x - 3)^2 + 0 \). Here \(a = 2\), \(h = 3\), \(k = 0\). Vertex: \((3, 0)\), a minimum (since \(a > 0\)).

B - X-Intercepts of a Quadratic Function in Vertex Form

The x-intercepts are the real solutions (if any) of the equation: \[ a(x - h)^2 + k = 0 \] 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 \(-\frac{k}{a} \ge 0\) (i.e., \(-\frac{k}{a}\) is non-negative).

Example: Finding X-Intercepts

Find the x-intercepts for each function.

  1. \( f(x) = -2(x - 3)^2 + 2 \)
  2. \( g(x) = -(x + 2)^2 \)
  3. \( h(x) = 4(x - 1)^2 + 5 \)

Solution:

  1. 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 \((2, 0)\).
  2. Solve \(-(x + 2)^2 = 0\):
    \[ x + 2 = 0 \Rightarrow x = -2 \] One x-intercept at \((-2, 0)\).
  3. Solve \(4(x - 1)^2 + 5 = 0\):
    \[ 4(x - 1)^2 = -5 \Rightarrow (x - 1)^2 = -\frac{5}{4} \] No real solution (negative right-hand side). Hence, no x-intercepts.

C - 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.

Example: Conversion

Rewrite \( f(x) = -(x - 2)^2 - 4 \) into general form.

Solution:

\[ \begin{aligned} f(x) &= -(x - 2)^2 - 4 \\ &= -(x^2 - 4x + 4) - 4 \\ &= -x^2 + 4x - 4 - 4 \\ &= -x^2 + 4x - 8 \end{aligned} \] So, \( a = -1, b = 4, c = -8 \) in general form.

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