4. The slope of the line 2x 7y = 8 is (a) 2/7 (b) 7/2 (c) 2 (d) 2/7 (e) None of these.

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1 Mat 11. Test Form N Fall 016 Name. Instructions. Te first eleven problems are wort points eac. Te last six problems are wort 5 points eac. For te last six problems, you must use relevant metods of algebra in your solutions, and sow all appropriate work in order to receive full credit. You may use a scientific nongraping calculator. Please do your best! 1. Given te circle (x ) + (y + 6) = 5. (a) Its center is: (b) Its radius is. Circle letter tat best describes te symmetry of te grap of 3 y = 9y 4 + x. (a) Te grap of 3 y = 9y 4 + x is symmetric about te x-axis because replacing x wit x leaves te equation uncanged. (b) Te grap of 3 y = 9y 4 + x is symmetric about te x-axis because replacing y wit y leaves te equation uncanged. (c) Te grap of 3 y = 9y 4 + x is symmetric about te y-axis because replacing x wit x leaves te equation uncanged. (d) Te grap of 3 y = 9y 4 + x is symmetric about te y-axis because replacing y wit y leaves te equation uncanged. (e) Te grap of 3 y = 9y 4 + x is symmetric about te origin because replacing (x, y) wit ( x, y) leaves te equation uncanged. 3. Te range, in interval notation, of te quadratic function f(x) = 3(x + 3) + 7 is (a) [7, ) (b) (, 7] (c) [ 7, ) (d) (, 7] (e) None of tese 4. Te slope of te line x 7y = 8 is (a) /7 (b) 7/ (c) (d) /7 (e) None of tese. 5. Te graps of y = x and of anoter function are given below. Circle te letter tat gives te equation of te oter grap. (a) y = x (b) y = x 3 5 (c) y = x (d) y = x 5 3 (e) None of tese y = x 6. A polynomial P wit integer coefficients satisfies P () = 18 and P ( 3) = 0. Wic of te following must be true concerning P (x)? Circle all coices tat apply. (a) (x 3) is a factor of P (x). (b) (x + 3) is a factor of P (x). (c) Te remainder of P (x) (x + ) is 18. (d) Te remainder of P (x) (x ) is 18. (e) Te remainder of P (x) (x 18) is. y x

2 7. Let P (x) = 3x x x 3x +. Circle letter tat best describes far rigt and far left beavior of P. (a) Up to far rigt and up to far left (b) Up to far rigt and down to far left (c) Down to far rigt and down to far left (d) Down to far rigt and up to far left (e) None of te above. 8. Let P (x) = (x 3) 8 (x + 5) 3. Wic of te following are true? Circle all coices tat apply. (a) Te grap of P (x) crosses te x-axis at (3, 0) (b) Te grap of P (x) crosses te x-axis at ( 5, 0) (c) Te grap of P (x) crosses te x-axis at (, 0). (d) Te grap of P (x) intersects but does not cross te x-axis at (3, 0). (e) Te grap of P (x) intersects but does not cross te x-axis at ( 5, 0). 9. According to te rational zero teorem, wic of te following are possible rational zeros of Q(x) = 13x 7 5x x + 7x 14? Circle all coices tat apply. (a) 1 13 (b) (c) (d) 1 14 (e) Let P (x) = x 3 + 3x 5. Use te Intermediate Value Teorem to determine weter P (x) as a zero between 0 and. (a) Because P (0) and P () ave opposite signs, we know tat P as at least one real zero between 0 and. (b) Because P (0) and P () ave opposite signs, we do not know if P as at least one real zero between 0 and. (c) Because P (0) and P () ave te same sign, we know tat P as at least one real zero between 0 and. (d) Because P (0) and P () ave te same sign, we do not know if P as at least one real zero between 0 and. (e) None of te above. 11. Find te orizontal asymptote(s) of f(x) = 3 x + x 4. Circle letter of best response. 5x 4 + x 3 1 (a) Te orizontal asymptote is y = 1/5. (b) Te orizontal asymptote is y = 3. (c) Te orizontal asymptote is y = 3/5. (d) Te orizontal asymptote is y = 0. (e) None of te above.

3 1. Julie opened a lemonade stand and found tat daily er profit is a linear function of te number of cups of lemonade sold. Wen se sells 360 cups of lemonade, se makes $4 and wen se sells 40 cups of lemonade, se makes $63. (a) Find te profit function. (b) How many cups of lemonade does Julie need to sell to make $16 in a day? Solution: (a) Let x be te number of cups sold, ten P (x) = mx + b were m = = 1 60 = 0.35 Tus P (x) =.0.35x + b, and so 4 = (360)(0.35) + b wic means b = 4 16 = 84. Hence P (x) = 0.35x 84. (b) To make $16, we solve 0.35x 84 = 16, and so x = ( )/(0.35) = 600 cups. 13. (a) Find te coordinates of te vertex of te grap of te quadratic function f defined by f(x) = 5x + 5x + 6 (b) Write te quadratic function f in standard form. Solution: (a) Te vertex for a function f(x) = ax + bx + c is (, k) were = b and a 4ac b k = f() =. In tis case f(x) = 5x + 5x + 6 and so a = 5, b = 5 and c = 6. 4a

4 Terefore, = 5 ( 5) = 1. Ten plugging tis value into f or using te vertex formula for k we find k = f ( ) 1 = 5 ( ) In eiter case, k = 9 4 ; terefore, te vertex is ( 1, 9 4 ) ( ) or k = (b) Standard form is f(x) = a(x ) + k were (, k) is te vertex. Terefore, ( f(x) = 5 x 1 ) Let f(x) = 7x 3x + 4. Find and simplify te difference quotient f(x + ) f(x). Solution: For f(x) = 7x 3x + 4, te difference quotient is f(x + ) f(x) = 7(x + ) 3(x + ) + 4 (7x 3x + 4) = 7(x + x + ) 3x x + 3x 4 = 14x = (14x + 7 3) = 14x + 7 3

5 15. Use syntetic division to find (x 4 6x 6x + 7) (x 3) Solution: Te syntetic division wit c = 3 is as follows Terefore, te answer is: x 3 + 3x + 3x x Find all zeros of te polynomial P defined by given tat x = i is a zero of P. P (x) = x 4 7x 3 + 5x 8x + 4 Solution: Because complex (non-real) zeros of polynomials wit real coefficients come in conjugate pairs, we know tat bot i and i are zeros of P (x). We now use syntetic division divide te factors wit tese zeros out of P (x). i i 4 14i 8 + i i 1 14i i 0 Now use syntetic division again to divide out te factor (x + i). i i 1 14i i i 14i i Tis means P (x) = (x i)(x + i)(x + 5)(x 7x + 1). To find te remaining two zeros of P, we solve solve x 7x + 1 = 0. According to te quadratic formula, te solutions are x = 7 ± (7) 4(1)(1) = 7 ± 45 Tus, te zeros of P are i, i, 7 45 and

6 17. Let f(x) = x 9x + 0 x 7x + 1. (a) State te domain of f, and simplify f if possible. (b) Find equations for te vertical asymptote(s) for te grap of f, and determine te beavior of f just to te rigt and just to te left of eac vertical asymptote. (c) Find all values of c for wic tere is a ole in te grap of f above x = c. Solution: (a) Te domain of f is {x : x 3, x 4}, and we simplify f as f(x) = (x 5)(x 4) (x 3)(x 4) = x 5 x 3 x 3, x 4 (b) Te zeros in te denominator of te simplified rational function provide te locations of te vertical asymptotes. Terefore, te grap of f as vertical asymptote x = 3. (c) For numbers x sligtly bigger tan 3 we ave f(x) = (x 5) (x 3) = a very large negative number. a small positive number Terefore, f decreases witout bound just to te rigt of te asymptote. For numbers x sligtly smaller tan 3 we ave f(x) = (x 5) (x 3) a small negative number = a very large positive number. Terefore, f increases witout bound just to te left of te asymptote. A table of values is as follows x (x 5) (x 3) (d) Tere would be a ole in te grap above x = 4 since 4 is not in te domain of f, but tere is no vertical asymptote at x = 4.