Precalculus Test 2 Practice Questions Page Note: You can expect oter types of questions on te test tan te ones presented ere! Questions Example. Find te vertex of te quadratic f(x) = 4x 2 x. Example 2. Given f(x) = x 3 f(x + ) f(x) +, simplify te quantity substituting zero for will not yield te indeterminant form 0 0. as muc as possible. You sould simplify until Example 3. Find te remainder r(x) wen g(x) = 4x 3 2x + 3 is divided by d(x) = 2x 8 using long division of polynomials. Example 4. Sketc te polynomial f(x) = (2x ) 3 (2 x) 2 by and. Sow all your work. Example 5. For te function g(x) given below, determine wat monomial te function approaces for large x. Ten, evaluate lim g(x) and lim g(x). Does te function g(x) ave any orizontal asymptotes? x x g(x) = ( x4 + 24x 78)( 2x + ) 3x 3 99 Example 6. Sketc te rational function (x) = 2(x 2 by and 4) (find x-intercepts, vertical asymptotes, slant or orizontal asymptotes, and end beaviour). Example 7. Solve te inequality and. Sow your work. Example 8. Solve te inequality Sow all your work. Example 9. Solve x 2 ( 4) 3x x + + 5 x 2 = 5 x 2 x 2 0 by constructing a sign cart, or drawing an appropriate sketc by x + 2 by constructing a sign cart, or drawing an appropriate sketc by and. x2 for x. Example 0. Te volume of an enclosed gas (at a constant temperature) varies inversely as te pressure. Tis means te relationsip between volume V and pressure P can be written as V = k P (Boyle s Law) were k is te proportionality constant. If te pressure of a 3.46 L sample of neon gas at 302 K is 0.926 atm, wat would te volume be at a pressure of.452 atm if te temperature does not cange? Since tis problem is from cemistry, you can ave a solution tat uses decimals rater tan fractions.
Precalculus Test 2 Practice Questions Page 2 Solutions Example. Find te vertex of te quadratic f(x) = 4x 2 x. Solution using completing te square: f(x) = 4x 2 x ( ( ) ) = 4 x 2 + x + 3 4 ( ( ) ( ) 2 ( ) ) 2 = 4 x 2 + x + 4 28 28 ( ( ) ( ) 2 ( ) ) 2 = 4 x 2 + x + 4 28 28 ( [ ( )] 2 ( ) ) 2 = 4 x + 28 28 ( [ = 4 x ] 2 ( ) ) 2 28 28 [ = 4 x ] 2 ( ) 2 4 28 28 [ = 4 x ] 2 28 56 Te vertex is (, k) = ( 28, ). 56 Anoter way to solve tis would be to recognize tat te x coordinate of te vertex is always centered between te zeros (even if te zeros are not real numbers!). Zeros: 4x 2 x = 0 x(4x ) = 0 x = 0 or x = /4 So te x coordinate of te vertex is 2 ( ) 0 + 4 = 28. Te y coordinate of te vertex is f(/28) = 4(/28) 2 (/28) = /56. ( Te vertex is (, k) = 28, ). 56
Precalculus Test 2 Practice Questions Page 3 Example 2. Given f(x) = x 3 f(x + ) f(x) +, simplify te quantity substituting zero for will not yield te indeterminant form 0 0. as muc as possible. You sould simplify until f(x + ) f(x) Average Rate of Cange = = (x + )3 + (x) 3 = x3 + 3x 2 + 3x 2 + 3 x 3 = 3x2 + 3x 2 + 3 = (3x2 + 3x + 2 ) = 3x 2 + 3x + 2 Example 3. Find te remainder r(x) wen g(x) = 4x 3 2x + 3 is divided by d(x) = 2x 8 using long division of polynomials.
Precalculus Test 2 Practice Questions Page 4 Example 4. Sketc te polynomial f(x) = (2x ) 3 (2 x) 2 by and. Sow all your work. Example 5. For te function g(x) given below, determine wat monomial te function approaces for large x. Ten, evaluate lim g(x) and lim g(x). Does te function g(x) ave any orizontal asymptotes? x x g(x) = ( x4 + 24x 78)( 2x + ) 3x 3 99 For end beaviour, we look at te leading terms in eac factor, since te leading terms will dominate for large x : g(x) = ( x4 + 24x 78)( 2x + ) 3x 3 99 ( x4 )( 2x) 3x 3 = 2 3 x2 for x large. Terefore, te end beaviour can be described as: Te function does not ave any orizontal asymptotes. lim f(x) = x lim f(x) =. x
Precalculus Test 2 Practice Questions Page 5 Example 6. Sketc te rational function (x) = 2(x 2 by and 4) (find x-intercepts, vertical asymptotes, slant or orizontal asymptotes, and end beaviour). Te numerator is already factored. Te denominator factors as 2x 2 8 = 2(x 2 4) = 2(x + 2)(x 2). Terefore, (x) = 2x 2 8 = (x + 6) 3 2(x + 2)(x 2). Te zero of te numerator is x = 6, wic is multiplicity 3 (odd), so te function will cross te x-axis ere. Since te multiplicity is greater tan 2, te grap will be flat (orizontal) near x = 6. Te denominator is zero wen x = 2 and x = 2, so tese are vertical asymptotes. Since te multiplicity of tese points is odd, te function will cange sign at x = 2 and x = 2. For end beaviour, we can look at wat appens for x very large: (x) = 2x 2 8 (x)3 2x 2 = x 2. Tis means tat for x very large te function (x) will approac te straigt line y = x/2. Tis is a slant asymptote. We ave enoug information to plot te function. Te dased line in te plot is te slant asymptote y = x/2. Example 7. Solve te inequality and. Sow your work. x 2 ( 4) I will solve tis using a sign cart, and examining te sign of te factors. Te numerator of te function f(x) = Te denominator of f is zero wen x = 5. x 2 ( 4) 0 by constructing a sign cart, or drawing an appropriate sketc by is zero wen x = 5/4, x = 2. (+) ( ) 0 ( ) ( ) 0 + ( ) ( ) + ( ) (+) x negative 5 4 positive 2 positive 5 negative From te sign cart, we see tat not defined tere. x 2 ( 4) 0 if x (, 5/4] (5, ). We exclude x = 5, since te function is
Precalculus Test 2 Practice Questions Page 6 Example 8. Solve te inequality Sow all your work. x + 2 by constructing a sign cart, or drawing an appropriate sketc by and. x2 We need to write tis as a single rational function, rater tan as a sum of rational functions, before we can construct our sign cart or a sketc. x + 2 + x 2 0 ( ) x 2 x + 2 x 2 + ( ) x + 2 x 2 0 x + 2 x 2 + x + 2 (x + 2)(x 2 ) Let s construct a sign cart. 0 Te quadratic in te numerator as no real roots. Te denominator is zero if x = 2, 0. Tese are te possible values were te function will cange sign. (+) ( )(+) (+) (+)(+) (+) (+)(+) x negative 2 positive 0 positive From te sign cart, we see tat x + 2 if x (, 2). x2 3x Example 9. Solve x + + 5 x 2 = 5 x 2 for x. x 2 3x x + + 5 x 2 = 5 x 2 x 2 3x x + + 5 x 2 = 5 (x 2)(x + ) [ 3x (x 2)(x + ) x + + 5 ] x 2 = 5 (x 2)(x + ) 3x(x 2) + 5(x + ) = 5 3x 2 6x + 5x + 5 5 = 0 3x 2 x 0 = 0 x 2, x 2, x 2, (3x + 5)(x 2) = 0 x 2, multiply by LCD (x 2)(x + ) Te solution to te original rational equation are x = 5/3. Te x = 2 is an extraneous solution.
Precalculus Test 2 Practice Questions Page 7 Example 0. Te volume of an enclosed gas (at a constant temperature) varies inversely as te pressure. Tis means te relationsip between volume V and pressure P can be written as V = k P (Boyle s Law) were k is te proportionality constant. If te pressure of a 3.46 L sample of neon gas at 302 K is 0.926 atm, wat would te volume be at a pressure of.452 atm if te temperature does not cange? Since tis problem is from cemistry, you can ave a solution tat uses decimals rater tan fractions. Te relation we ave is V = k P We can use te first data point to determine te proportionality constant k: k = V P = (3.46)(0.926) = 3.20396. Te relationsip is terefore V = 3.20396 P and we ave V = 3.20396.452 = 2.20658. So te volume would be 2.20658 L.