# Honors Algebra 2: Unit 5 Extra Review Problems

• Class: Honors Algebra 2
• Author: Peter Atlas
• Text: Algebra and Trigonometry: Structure and Method, Brown

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1. Consider the function $$\displaystyle f(x) = \frac{x^2 - x - 6}{x^3 + 2x^2 - x - 2}$$
1. Find the coordinate(s) of any hole(s). If there are none, so state.
2. Solution Factoring, we get $$\displaystyle f(x) = \frac{(x + 2)(x - 3)}{(x + 2)(x + 1)(x - 1)}.$$ The factors of $$(x + 2)$$ reduce, causing a hole where that factor would equal zero, at $$x = -2$$. The reduced fraction is $$\displaystyle f(x) = \frac{(x - 3)}{(x + 1)(x - 1)}.$$ Plugging in x = -2 to the reduced form gives $$\displaystyle f(-2) = -\frac{5}{3}$$ So the hole occurs at $$\left( -2, -\frac{5}{3} \right)$$
3. Find the equation of any vertical asymptote(s). If there are none, so state.
4. Solution The remaining factors in the denominator are $$x + 1$$ and $$x - 1$$. These are where the vertical asymptotes would be -- at the values of $$x$$ that make those factors zero. $$x = \pm 1$$
5. Find the equation of any horizontal asymptote(s). If there are none, so state.
6. Solution The dominant terms are $$\displaystyle \frac{x}{x^2}$$. That reduces to $$\displaystyle \frac{1}{x}$$. As x approaches infinity, $$\displaystyle \frac{1}{x}$$ approaches 0, so the horizontal asymptote is $$y = 0.$$
7. Find the coordinates of the y-intercept, if there is one.
8. Solution Plug in 0 for $$x. (0, 3)$$
9. Find the coordinates of the x-intercept(s), if there are any. If there are none, so state.
10. Solution Plug in 0 for $$y. (3, 0)$$
11. Sketch the graph of y = f(x)
12. Solution
Note: to graph this, I plotted the intercepts, and sketched the asymptotes. Then, for each region, I picked a point in the region (eg. to the left of the hole, pick $$x = -10$$. Between the y-intercept and the left vertical asymptote, pick $$\displaystyle x = -\frac{1}{2}.$$ Plug the x-values in to get y-values, and that will help you figure out whether the graph is above or below the horizontal asymptotes.
2. Solve: $$\displaystyle \frac{x + 2}{x - 1} - \frac{x - 4}{x + 3} = \frac{-28}{x^2 + 2x - 3}$$
3. Solution The empty set. The only answer would be $$x = -3$$. but -3 is excluded from the domain of the question.
4. Simplify: $$\displaystyle 8^{x + 1} \cdot 4^{2x} \cdot 2^{3x + 3}$$
5. Solution This is a multiplication problem. Get everything in terms of a single base: $$\left( 2^3 \right) ^{x + 1} \cdot \left( \left( 2 \right) ^2 \right)^{2x} \cdot 2^{3x + 3} = 2^{3x + 3 + 4x + 3x + 3} = 2^{10x + 6}$$
6. Simplify: $$\displaystyle 8^{x + 1} + 4^{2x} + 2^{3x + 3}$$
7. Solution Now we can only add if we have like terms -- so we need like bases and like powers. $$\displaystyle \left( 2^3 \right) ^{x + 1} + \left( 2 ^2 \right) ^{2x} + 2^{3x + 3} = 2^{3x + 3} + 2^{4x} + 2^{3x + 3}.$$ Two terms are like, one is not. When combining like terms, only the coefficient changes: $$\displaystyle 2 \cdot 2^{3x + 3} + 2^{4x}.$$ Now that first term can be tweaked when you realize that 2 is the same as $$2^1$$. The final answer is $$2^{3x + 4} + 2^{4x}.$$
8. Simplify: $$(a^{-1} + b^{-1})(a + b)^{-1}$$
9. Solution Since the second factor: $$(a + b)^{-1}$$ is being multiplied by the first factor, it may go down to the denominator of the resulting compound fraction. But since in the first factor has a sum in it, the individual terms being raised to negative powers: $$a^{-1}$$ and $$b^{-1}$$ can't. So the resulting compound fraction that gets rid of all the negative exponents is $$\displaystyle \frac{\frac{1}{a} + \frac{1}{b}}{( a + b)}$$. Multiplying numerator and denominator by ab to collapse the fraction, we get: $$\displaystyle \frac{b + a}{(ab)(a + b)}$$ which reduces to $$\displaystyle \frac{1}{ab}.$$ And don't forget the domain restriction that gets lost in the reduction: $$a \neq - b$$.