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Magazine Chapter 4: Problem 6
For each of the following, check for discontinuities and then use at least two other tests to make a rough sketch of the curve. Verify using a calculator. a. \(y=\frac{x-3}{x+5}\) b. \(f(x)=\frac{5}{(x+2)^{2}}\) c. \(g(t)=\frac{t^{2}-2 t-15}{t-5}\) d. \(y=\frac{(2+x)(3-2 x)}{\left(x^{2}-3 x\right)}\)
Short Answer
Expert verified
Discontinuities: a) x=-5, b) x=-2, c) t=5, d) x=0,3. Transformation reveals behavior at these points and helps sketch curves.
Step by step solution
01
Identify Discontinuities for y=\frac{x-3}{x+5}
The function is undefined where the denominator equals zero. Set the denominator to 0: \(x+5=0\). Solving gives \(x=-5\). Thus, there is a discontinuity at \(x=-5\).
02
Find Additional Features for y=\frac{x-3}{x+5}
Look for horizontal or vertical asymptotes. The vertical asymptote is at \(x=-5\) from our discontinuity. To check for horizontal asymptotes, compare the degrees of the numerator and denominator. Since both are linear, the horizontal asymptote is \(y=1\). No intercepts are found since \(x=3\) gives \(y=0\).
03
Identify Discontinuities for f(x)=\frac{5}{(x+2)^{2}}
The function is undefined where the denominator equals zero. Set \((x+2)^2=0\), solving gives \(x=-2\). There is a discontinuity at \(x=-2\).
04
Find Additional Features for f(x)=\frac{5}{(x+2)^{2}}
Check for vertical asymptotes. A vertical asymptote is at \(x=-2\). Look for horizontal asymptotes by considering limits as \(x\to\pm\, \infty\), which is \(y=0\). No x-intercepts are present since the numerator is a constant.
05
Identify Discontinuities for g(t)=\frac{t^{2}-2t-15}{t-5}
The function is undefined where \(t-5=0\). Thus, \(t=5\) is a discontinuity. Factor the numerator: \((t-5)(t+3)\). This simplifies the function to \(t+3\) for \(teq5\). Hence, \(t=5\) is a removable discontinuity (hole).
06
Find Additional Features for g(t)=\frac{t^{2}-2t-15}{t-5}
After simplification, \(g(t)=t+3\) for \(teq5\). This is a straight line with slope 1 passing through \((5,8)\) as a hole. The y-intercept is \(g(0)=3\).
07
Identify Discontinuities for y=\frac{(2+x)(3-2x)}{x^2-3x}
Find x-values where the denominator is 0: \(x(x-3)=0\). This gives discontinuities at \(x=0\) and \(x=3\).
08
Find Additional Features for y=\frac{(2+x)(3-2x)}{x^2-3x}
Identify vertical asymptotes: \(x=0\) and \(x=3\). Simplify the expression by cancelling common factors if any. Find horizontal asymptote considering limits as \(x\to\mp \infty\), since numerator and denominator have the same degree: \(y=-2\). The intercepts can be found by solving \(y=0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Asymptotes
Understanding asymptotes can make analyzing functions much simpler. Asymptotes are lines that a graph approaches but never actually touches. These lines can either be vertical or horizontal, depending on the behavior of the function.
- **Vertical asymptotes** occur when the denominator of a rational function becomes zero, leading to undefined points. For example, in the function \( y = \frac{x-3}{x+5} \), a vertical asymptote occurs at \( x = -5 \), where the denominator becomes zero.
- **Horizontal asymptotes** appear when we compare the degrees of the polynomial in the numerator and denominator. If both have the same degree, as seen in \( y = \frac{x-3}{x+5} \), the horizontal asymptote is the ratio of their leading coefficients, which is \( y = 1 \).
Removable Discontinuity
Removable discontinuities, often known as "holes," occur in rational functions when a common factor exists in both the numerator and denominator. Such points aren't part of the graph, but the function can be defined or tweaked to "fill in" the gap.
Take the function \( g(t) = \frac{t^{2}-2t-15}{t-5} \) as an instance. Factoring the numerator gives \((t-5)(t+3)\), which cancels with the denominator \( t-5 \), leading to the simplified function \( g(t) = t+3 \) for \( t eq 5 \). Though \( g(t) \) simplifies to a linear function, it's undefined at \( t = 5 \). This creates a hole located at \( (5, 8) \) after simplification. Such discontinuities can be "filled" by extending the function definition to include those specific points.
Take the function \( g(t) = \frac{t^{2}-2t-15}{t-5} \) as an instance. Factoring the numerator gives \((t-5)(t+3)\), which cancels with the denominator \( t-5 \), leading to the simplified function \( g(t) = t+3 \) for \( t eq 5 \). Though \( g(t) \) simplifies to a linear function, it's undefined at \( t = 5 \). This creates a hole located at \( (5, 8) \) after simplification. Such discontinuities can be "filled" by extending the function definition to include those specific points.
Rational Functions
Rational functions are ratios of polynomials, written as \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. These functions often have interesting characteristics due to division, including discontinuities.
Consider the function \( f(x) = \frac{5}{(x+2)^2} \). It is undefined where \( (x+2)^2 = 0 \), which happens at \( x = -2 \). Uniquely, rational functions can show both removable and non-removable discontinuities, alongside vertical and horizontal asymptotes. Since the degree of the numerator is less than the degree of the denominator, this function has a horizontal asymptote at \( y = 0 \).
Examining rational functions thoroughly can reveal vertical asymptotes, removable discontinuities, and the end behaviors, all of which are crucial for sketching an accurate graph.
Consider the function \( f(x) = \frac{5}{(x+2)^2} \). It is undefined where \( (x+2)^2 = 0 \), which happens at \( x = -2 \). Uniquely, rational functions can show both removable and non-removable discontinuities, alongside vertical and horizontal asymptotes. Since the degree of the numerator is less than the degree of the denominator, this function has a horizontal asymptote at \( y = 0 \).
Examining rational functions thoroughly can reveal vertical asymptotes, removable discontinuities, and the end behaviors, all of which are crucial for sketching an accurate graph.
Slope in Linear Functions
The slope of a line describes its steepness and direction. For linear functions, expressed in the form \( y = mx + b \), \( m \) represents the slope.
In the expression \( g(t) = t+3 \), derived from simplifying \( g(t) = \frac{t^{2}-2t-15}{t-5} \), the slope \( m = 1 \). This means as you move 1 unit right on the graph, \( g(t) \) increases by 1 unit. The y-intercept, \( b = 3 \), tells us that the line crosses the y-axis at \( (0, 3) \).
Understanding slopes in simplified versions of rational functions is invaluable for identifying lines and predicting graph behavior even when holes suggest potential discontinuities. Linear function properties apply generally, but factoring and simplifying reveal these aspects in more complex equations.
In the expression \( g(t) = t+3 \), derived from simplifying \( g(t) = \frac{t^{2}-2t-15}{t-5} \), the slope \( m = 1 \). This means as you move 1 unit right on the graph, \( g(t) \) increases by 1 unit. The y-intercept, \( b = 3 \), tells us that the line crosses the y-axis at \( (0, 3) \).
Understanding slopes in simplified versions of rational functions is invaluable for identifying lines and predicting graph behavior even when holes suggest potential discontinuities. Linear function properties apply generally, but factoring and simplifying reveal these aspects in more complex equations.
