Chapter 2: Problem 40
Find a polynomial (there are many) of minimum degree that has the given zeros. $$0,1,3,5,10$$
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Chapter 2: Problem 40
Find a polynomial (there are many) of minimum degree that has the given zeros. $$0,1,3,5,10$$
These are the key concepts you need to understand to accurately answer the question.
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Explain the mistake that is made. Determine whether a horizontal or a slant asymptote exists for the function \(f(x)=\frac{9-x^{2}}{x^{2}-1} .\) If one does, find it. Solution: Step 1: The degree of the numerator equals the degree of the denominator, so there is a horizontal asymptote. Step 2: The horizontal asymptote is the ratio of the lead $$\text { coefficients: } y=\frac{9}{1}=9$$ The horizontal asymptote is \(y=9\) The following is a correct graph of the function. (GRAPH CAN'T COPY) Note that there is no horizontal asymptote at \(y=9 .\) What went wrong?
In Exercises \(87-90\), determine whether each statement is true or false. A quadratic function must have an \(x\) -intercept.
In Exercises \(55-66\), find the quadratic function that has the given vertex and goes through the given point. vertex: (0,-2)\(\quad\) point: (3,10)
(a) Identify all asymptotes for each function. (b) Plot \(f(x)\) and \(g(x)\) in the same window. How does the end behavior of the function \(f\) differ from that of \(g ?\) (c) Plot \(g(x)\) and \(h(x)\) in the same window. How does the end behavior of \(g\) differ from that of \(h ?\) (d) Combine the two expressions into a single rational expression for the functions \(g\) and \(h .\) Does the strategy of finding horizontal and slant asymptotes agree with your findings in (b) and (c)? $$\begin{aligned} &f(x)=\frac{1}{x-3}, g(x)=2+\frac{1}{x-3}, \text { and }\\\ &h(x)=-3+\frac{1}{x-3} \end{aligned}$$
In Exercises \(87-90\), determine whether each statement is true or false. A quadratic function may have more than one \(y\) -intercept.
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