Problem 52 - Intermediate Algebra

Question

Graph each of the following parabolas, and use the TRACE feature to find whole number estimates of the vertex. Then either complete the square or use \(\left(-\frac{b}{2 a}, \frac{4 a c-b^{2}}{4 a}\right)\) to find the vertex. (a) \(f(x)=x^{2}-6 x+3\) (b) \(f(x)=x^{2}-18 x+66\) (c) \(f(x)=-x^{2}+8 x-3\) (d) \(f(x)=-x^{2}+24 x-129\) (e) \(f(x)=14 x^{2}-7 x+1\) (f) \(f(x)=-0.5 x^{2}+5 x-8.5\)

Step-by-Step Solution

Answer

Vertices: (3, -6), (9, -15), (4, 13), (12, 15), \(\left(\frac{1}{4}, \frac{-5}{28}\right)\), (5, 4).

Step 1: Identify Vertex Formula

We will use the formula for the vertex of a parabola given by \((-\frac{b}{2a}, \frac{4ac-b^2}{4a})\). For each quadratic function of the form \(f(x) = ax^2 + bx + c\), `a`, `b`, and `c` are coefficients that we will identify in each part.

Step 2: Calculate Vertex for Part (a)

For \(f(x) = x^2 - 6x + 3\), identify \(a = 1\), \(b = -6\), and \(c = 3\). The vertex \((h, k)\) can be found using:\[h = -\frac{-6}{2 \times 1} = 3\]\[k = \frac{4 \times 1 \times 3 - (-6)^2}{4 \times 1} = -6\]So, the vertex is \((3, -6)\).

Step 3: Calculate Vertex for Part (b)

For \(f(x) = x^2 - 18x + 66\), identify \(a = 1\), \(b = -18\), and \(c = 66\). The vertex is:\[h = -\frac{-18}{2 \times 1} = 9\]\[k = \frac{4 \times 1 \times 66 - (-18)^2}{4 \times 1} = -15\]Thus, the vertex is \((9, -15)\).

Step 4: Calculate Vertex for Part (c)

For \(f(x) = -x^2 + 8x - 3\), identify \(a = -1\), \(b = 8\), and \(c = -3\). The vertex is:\[h = -\frac{8}{2 \times -1} = 4\]\[k = \frac{4 \times -1 \times -3 - 8^2}{4 \times -1} = 13\]Hence, the vertex is \((4, 13)\).

Step 5: Calculate Vertex for Part (d)

For \(f(x) = -x^2 + 24x - 129\), identify \(a = -1\), \(b = 24\), and \(c = -129\). The vertex is:\[h = -\frac{24}{2 \times -1} = 12\]\[k = \frac{4 \times -1 \times -129 - 24^2}{4 \times -1} = 15\]The vertex is \((12, 15)\).

Step 6: Calculate Vertex for Part (e)

For \(f(x) = 14x^2 - 7x + 1\), identify \(a = 14\), \(b = -7\), and \(c = 1\). The vertex is:\[h = -\frac{-7}{2 \times 14} = \frac{1}{4}\]\[k = \frac{4 \times 14 \times 1 - (-7)^2}{4 \times 14} = \frac{-5}{28}\]Thus, the vertex is \(\left(\frac{1}{4}, \frac{-5}{28}\right)\).

Step 7: Calculate Vertex for Part (f)

For \(f(x) = -0.5x^2 + 5x - 8.5\), identify \(a = -0.5\), \(b = 5\), and \(c = -8.5\). The vertex is:\[h = -\frac{5}{2 \times -0.5} = 5\]\[k = \frac{4 \times -0.5 \times -8.5 - 5^2}{4 \times -0.5} = 4\]So, the vertex is \((5, 4)\).

Key Concepts

Quadratic Functions

A quadratic function is a type of polynomial where the highest exponent of the variable is 2, often written in the form \( f(x) = ax^2 + bx + c \). This is called the standard form, where \( a \), \( b \), and \( c \) are constants, and \( a \) cannot be zero. This kind of function describes a parabola when graphically depicted.
Quadratic functions have a unique feature, the vertex, which is the highest or lowest point of the parabola, depending on the sign of \( a \). A positive \( a \) means the parabola opens upwards, resembling a 'U' shape, whereas a negative \( a \) indicates it opens downwards.
The symmetry of quadratics is also interesting—every parabola is symmetrical around a vertical line that passes through its vertex. This line is called the axis of symmetry and it can be calculated using the formula \( x = -\frac{b}{2a} \). Understanding these aspects is crucial for solving and graphing quadratic functions.

Completing the Square

Completing the square is a method used to transform a quadratic equation into a perfect square trinomial. This technique simplifies the process of finding the vertex of a parabola, among other applications. Let's break down the method.

• Start with a quadratic in the form \( ax^2 + bx + c \).
• Move the constant term, \( c \), to the other side: \( ax^2 + bx = -c \).
• If \( a \) is not 1, factor it out from the left side, leaving \( x^2 + \frac{b}{a}x \).
• Take half of the \( b/a \) coefficient, square it, and add it inside the parenthesis on the left side. Remember to balance the equation by adding the same value to the right side.
  Now the equation represents a square, both sides can be simplified.

By manipulating it back into vertex form \( f(x) = a(x-h)^2 + k \), where \( (h, k) \) is the vertex, completing the square becomes a straightforward way to analyze and graph parabolas.

Graphing Parabolas

Graphing parabolas is an essential skill when working with quadratic functions. A parabola is the graph of a quadratic function, and its shape provides valuable information about the function's behavior.

• Identify the vertex using either the vertex formula or by completing the square. The vertex \( (h, k) \) is the central feature of the graph and is the point where the parabola changes direction.
• Determine if the parabola opens upwards or downwards by observing the sign of \( a \). Positive \( a \) means it opens upwards, while a negative \( a \) means it opens downwards.
• Find the axis of symmetry, a vertical line passing through the vertex. It divides the parabola into two mirror-image halves.
• Plot additional points on either side of the vertex to help in sketching a precise curve. Substitute values in the function to find corresponding \( y \)-values.

Graphing accurately involves a combination of these steps. Each element, like the vertex and symmetry, gives insight into the qualitative nature of the parabola, making analysis easier.