College Algebra

College Algebra—Exam 3 Show all your work for full credit, total 100 points Name: _________________________ Q1. (10 points) For the polynomial function 𝑔(π‘₯) = 3π‘₯ 7 βˆ’ π‘₯ 5 + 5π‘₯ 4 + π‘₯ 3 + 8 a) List the potential rational zeros. b) Use Descartes Rule of Signs to determine how many positive and negative real zeros each polynomial function may have. Do not attempt to find the zeros. Q2. (10 points) Find a polynomial with integer coefficients that satisfies the given condition. Q has degree 3 and zeros βˆ’3 and 1 + 𝑖 1 College Algebra—Exam 3 Show all your work for full credit, total 100 points Name: _________________________ Q3. (20 points) Given 𝑃(π‘₯) = π‘₯ 4 βˆ’ 2π‘₯ 3 βˆ’ 2π‘₯ 2 βˆ’ 2π‘₯ βˆ’ 3 a) Use synthetic division to find the quotient and remainder when 𝑃(π‘₯) is divided by π‘₯ βˆ’ 1 b) Is (π‘₯ βˆ’ 3) a factor of 𝑃(π‘₯)? c) Find all the real and complex zeros for 𝑃(π‘₯) d) Graph 𝑃(π‘₯) 2 College Algebra—Exam 3 Show all your work for full credit, total 100 points Name: _________________________ Q4. (10 points) Find the intercepts and asymptotes, and then sketch a graph of the rational function and state the domain and range. 6 π‘Ÿ(π‘₯ ) = π‘₯βˆ’2 3 College Algebra—Exam 3 Show all your work for full credit, total 100 points Name: _________________________ Q5. (10 points) Solve the inequality 2π‘₯ 2 > π‘₯ + 3 π‘₯+2 Q6. (10 points) Solve the inequality π‘₯ 2 βˆ’9 4 ≀0 College Algebra—Exam 3 Show all your work for full credit, total 100 points Name: _________________________ Q7. (10 points) Find the common factors of the numerator and denominator of the rational function. Then find the intercepts and asymptotes and sketch a graph. State the domain and range. π‘Ÿ(π‘₯) = π‘₯ 2 βˆ’ 2π‘₯ βˆ’ 3 π‘₯βˆ’3 5 College Algebra—Exam 3 Show all your work for full credit, total 100 points Name: _________________________ Q8. (20 points) Given π‘Ÿ(π‘₯) find its domain, label all intercepts, vertical asymptotes, horizontal asymptotes, and oblique asymptotes. Then graph it π‘₯2 + π‘₯ βˆ’ 2 (π‘₯ βˆ’ 1)(π‘₯ + 2) π‘Ÿ(π‘₯) = 2 = π‘₯ βˆ’ 2π‘₯ βˆ’ 3 (π‘₯ + 1)(π‘₯ βˆ’ 3) 6
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