Mini Exam 5
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The rational function f is given by $$f(x)=\frac{x^3+2x^2-5x}{x^2-x}$$. In the $$xy$$-plane, which of the following is the slope of a slant asymptote of the graph of $$f$$?⌄
Perform polynomial division.$$\frac{x^3+2x^2-5x}{x^2-x}$$Divide leading terms.$$\frac{x^3}{x^2}=x$$Multiply and subtract.$$x(x^2-x)=x^3-x^2$$$$2x^2-(-x^2)=3x^2$$Continue.$$\frac{3x^2}{x^2}=3$$So, the quotient is $$x+3$$The slope of the slant asymptote is the coefficient of $$x$$.Final answer: $$1$$
Let $$f(x)=\cos x$$. The function $$g$$ has the same amplitude as $$f$$, and the period of $$g$$ is $$\frac{1}{3}$$ of the period of $$f$$. Which of the following defines $$g$$ in terms of $$f$$?⌄
The period of $$\cos x$$ is $$2\pi$$.The new period is:$$\frac{1}{3}\cdot2\pi=\frac{2\pi}{3}$$Use$$\frac{2\pi}{b}=\frac{2\pi}{3}$$Solve.$$b=3$$So, the function becomes $$f(3x)$$Amplitude stays the same.Final answer: $$\text{C}$$
The binomial theorem can be used to expand an expression of the form $$(a+b)^n$$. Which of the following is equivalent to $$(x+2y)^5$$?⌄
Use binomial coefficients.$$\binom{5}{0}=1$$$$\binom{5}{1}=5$$$$\binom{5}{2}=10$$$$\binom{5}{3}=10$$$$\binom{5}{4}=5$$$$\binom{5}{5}=1$$Expand.$$x^5+5x^4(2y)+10x^3(2y)^2+10x^2(2y)^3+5x(2y)^4+(2y)^5$$Final answer: $$\text{C}$$
The polar function $$r=f(\theta)$$, where $$f(\theta)=2-3\cos(-\theta)$$, is graphed in the polar coordinate system. As $$\theta$$ varies from $$\frac{\pi}{2}$$ to $$\pi$$, how is the distance between the origin and the point $$(f(\theta),\theta)$$ changing?⌄
$$\cos(-\theta)=\cos(\theta)$$So$$f(\theta)=2-3\cos(\theta)$$The distance from the origin is $$|r|$$.For $$\frac{\pi}{2}\le\theta\le\pi$$, $$\cos(\theta)$$ decreases from $$0$$ to $$-1$$.Then $$-3\cos(\theta)$$ increases from $$0$$ to $$3$$.So, $$r=2-3\cos(\theta)$$ increases from $$2$$ to $$5$$.Since $$r$$ stays positive on this interval, the distance from the origin is increasing.Final answer: $$\text{C}$$
The function $$f$$ is given by $$f(x)=x-2$$, and the function $$g$$ is given by $$g(x)=(x-2)(2x+1)$$. Consider the rational function $$m(x)=\frac{f(x)}{g(x)}$$. Which of the following is true about holes in the graph of $$m$$ in the $$xy$$-plane?⌄
Write the rational function.$$m(x)=\frac{x-2}{(x-2)(2x+1)}$$Cancel the common factor.$$m(x)=\frac{1}{2x+1}$$The factor $$x-2$$ cancels completely.So, there is a hole at $$x=2$$The factor $$2x+1$$ does not cancel.So, there is not a hole at $$x=-\frac{1}{2}$$Final answer: $$\text{D}$$
Consider the functions $$g$$ and $$h$$ given by $$g(\theta)=\tan\theta$$ and $$h(\theta)=\sec\theta$$. Which of the following statements is true for both $$g$$ and $$h$$?⌄
$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$$$\sec\theta=\frac{1}{\cos\theta}$$Both functions are undefined when $$\cos\theta=0$$This happens at $$\theta=\frac{\pi}{2}+\pi n$$ where $$n$$ is an integer.So, both graphs have vertical asymptotes at those values.Final answer: $$\text{C}$$
Both nonzero functions $$f$$ and $$g$$ are invertible. The input values of $$f$$ are distances, in miles, and the outputs are times, in hours. The input values of $$g$$ are times, in hours, and the outputs are costs, in dollars. For which of the following are the input values costs, in dollars, and the output values distance, in miles?⌄
$$f$$ maps $$\text{distance}\to\text{time}$$So $$f^{-1}$$ maps $$\text{time}\to\text{distance}$$$$g$$ maps $$\text{time}\to\text{cost}$$So, $$g^{-1}$$ maps $$\text{cost}\to\text{time}$$Compose them in this order.$$\text{cost}\to\text{time}\to\text{distance}$$This is $$f^{-1}(g^{-1}(x))$$Final answer: $$\text{B}$$
A data set is modeled using both linear and exponential regressions. The residual plot for the linear model shows a clear curved pattern, while the residual plot for the exponential model shows points randomly scattered around zero. Which of the following statements is true?⌄
A good residual plot has points randomly scattered around zero. A curved residual pattern indicates the model is not appropriate. The linear model has the curved pattern. So, the linear model is not appropriate. The exponential model has random scatter around zero. So, the exponential model is more appropriate.Final answer: $$\text{D}$$