Honors Algebra 2: Unit 5 Extra Review Problems

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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 \( \left( -2, -\frac{5}{3} \right) \)
    2. Find the equation of any vertical asymptote(s). If there are none, so state.
      3. 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. \(x = \pm 1\)
    3. Find the equation of any horizontal asymptote(s). If there are none, so state.
      4. 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 \(y = 0.\)
    4. Find the coordinates of the y-intercept, if there is one.
      5. Solution Plug in 0 for \(x. (0, 3)\)
    5. Find the coordinates of the x-intercept(s), if there are any. If there are none, so state.
      6. Solution Plug in 0 for \(y. (3, 0)\)
    6. Sketch the graph of y = f(x)
      7. 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.
  3. Simplify: \(\displaystyle 8^{x + 1} \cdot 4^{2x} \cdot 2^{3x + 3}\)
    4. Solution This is a multiplication problem. Get everything in terms of a single base: \( \left( 2^3 \right) ^{x + 1} \cdot \left( \left( 2 \right) ^2 \right)^{2x} \cdot 2^{3x + 3} = 2^{3x + 3 + 4x + 3x + 3} = 2^{10x + 6}\)
  4. Simplify: \(\displaystyle 8^{x + 1} + 4^{2x} + 2^{3x + 3}\)
    5. 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 \(2^{3x + 4} + 2^{4x}.\)
  5. Simplify: \( (a^{-1} + b^{-1})(a + b)^{-1}\)
    6. 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: \( a \neq - b\).