so f can now be rewritten as a product of g(x) = x 2 and the previous piecewisedefined

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1 Version PREVIEW HW 01 hoffman 575) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. FuncPcwise01a points By removing absolute values, express fx) = x 4 x + as a piecewise-defined function. where the point x = has been excluded from the domain because the quotient is not defined at that point. On the other hand, x 4 = x )x + ), so f can now be rewritten as a product x + ) fx) = gx) x + of gx) = x and the previous piecewisedefined function. Consequently, 1. fx) = { x +, x >, x + ), x <. fx) = { x, x >, x, x <.. fx) = 3. fx) = 4. fx) = 5. fx) = 6. fx) = { x +, x >, x + ), x <. { x, x >, x, x <. { x + ), x >, x +, x <. { x, x >, x, x <. correct { x, x >, x, x <. keywords: piecewise-defined function, absolute value, Gate points If the graph of f is Since a = { a,, a 0, a, a < 0, which of the following could be the graph of y = fx )? we see that x + = { x +, x, x + ), x <. 1. In particular, therefore, x + x + = { 1, x >, 1, x <,

2 Version PREVIEW HW 01 hoffman 575). correct 3. keywords: graph transformation, shift, horizontal shift, vertical shift, CalC1b13a points For a child between the ages of 6 and 16, its height y in inches) is approximately a linear function of age t in years). Joe Smith s son John was 6 inches tall at age 6 and 40 inches tall at age 10. Estimate John s height at age height = 49 inches. height = 45 inches 4. The graph of y = fx ) is obtained from that of y = fx) by translating to the right by units and then shifting vertically downwards by units. Consequently, the graph of is y = fx ) 3. height = 47 inches correct 4. height = 46 inches 5. height = 48 inches Since there is a linear relationship between height and age, the given data for John specifies two points on the line, namely the points 6, 6) and 10, 40). The two-point formula y y 1 = y y 1 t t 1 t t 1 ) can now be used to determine the linear equation relating John s age to his height. A computation shows that Thus y y 1 t t 1 = 7. y = 7 t + 5. At age 1, therefore it is estimated that John s height = 47 inches. Intercept05a points

3 Version PREVIEW HW 01 hoffman 575) 3 The manager of a weekend flea market knows from past experience that if he charges $x dollars for a rental space at the flea market, then the number y of spaces he can rent is given by the linear function y = fx) = 30 x, x 15. Which of the following statements are true? A. The x-intercept of the graph of f is 15 and a $15 rental fee results in no spaces being rented. B. The slope of the graph of f is which means that for each increase of $1 for a rental space, the number of spaces rented decreases by. C. The y-intercept of the graph of f is 15 and this is the number of spaces that would be occupied if there were no charge for each space. 1. A only. A and C only 3. A and B only correct 4. all of them 5. B and C only 6. none of them 7. C only 8. B only A. TRUE: the x-intercept of the graph of y = fx) is the solution of Thus y = fx) = 0. fx) = 30 x = 0 = x = 15. B. TRUE: in words, slope = rise run, so if the slope =, then the rise = whenever the run = 1. Thus the number of spaces decreases by for every $1 increase in rental charge. C. FALSE: the y-intercept of the graph of y = fx) is the value of y when x = 0. Thus x = 0 = y = f0) = keywords: CalCappC8t points Determine the vertex of the parabola whose x-intercepts are at 8, 0), 4, 0) and whose y-intercept is at 0, 4). 1. vertex =. vertex = 3. vertex = 4. vertex = 5. vertex =, 9 ) correct 6, 9 ) 6, 9 ) 6, 9 ), 9 ) If a parabola has x-intercepts are x = 8 and x = 4, then its equation must have the form y = Ax + 8)x 4) for some constant A. This constant is determined by the y-intercept 4 of the parabola, for y0) = 4 = 3A = 4. Thus the equation of the parabola is ) y = 1 x + 8)x 4). 8

4 Version PREVIEW HW 01 hoffman 575) 4 To determine the vertex of the parabola we complete the square in ). This shows that y = 1 8 x + 4x 3) = 1 8 x + ) 4 3) = 1 8 x + ) 36). Hence the vrtex of the parabola occurs at, 9 ). Graph10exam points For which of the following equations is its graph a straight line? 1. x y) = 0 correct. y x = 0 3. y + x = 0 4. x + y = 0 5. x y = 0 6. y + x = 0 7. y x = 0 Let s look at all the equations: i) the only solution of x y) = 0 is y x = 0. Its graph is a single straight line. ii) Since the solutions of x y = x y)x + y), x y = 0 are y x = 0 and y + x = 0. Its graphs are a pair of straight lines. iii) The only solution of x + y = 0 is x = 0 = y, so its graph is the single point 0, 0). iv) The graph of y x = 0 is a parabola symmetric about the y-axis and opening upwards. v) The graph of y + x = 0 is a parabola symmetric about the y-axis and opening downwards. vi) The graph of y x = 0 is a parabola symmetric about the x-axis and opening to the right. vii) The graph of y + x = 0 is a parabola symmetric about the x-axis and opening to the left. Consequently the equation whose graph is a straight line is keywords: x y) = 0. QuadAnalz05b points In which quadrant does the vertex of the parabola lie? y = 5 x x 1. Quadrant III correct

5 Version PREVIEW HW 01 hoffman 575) 5. Quadrant II 3. Quadrant I 4. Quadrant IV The vertex of the parabola y = b x a) is the point a, b), and so we can determine the quadrant in which this vertex lies by looking at the signs of a and b. To express in the form y = 5 x x B. If both x-intercepts are positive and the y-intercept is negative, then the graph opens downwards. 1. B only. both of them correct 3. neither of them 4. A only A. TRUE: if there are no x-intercepts then the graph does not cross the x-axis, and so if the y-intercept is positive the graph must look like y y = b x a) we must first complete the square: x x 5 = x + x) 5 = x + x + 1 1) 5 x = x + 1) = 4 x + 1). Consequently, the vertex of the parabola is the point 1, 4) which lies in Quadrant III. or some translate of it. Thus the graph must open upwards, not downwards. B. TRUE: if both x-intercepts are positive then the graph crosses the x-axis to the right of the origin, and so if the y-intercept is negative the graph must look like keywords: parabola, vertex, quadrant, complete square, QuadGraph01a points y x Which, if any, of the following statements are true for the graph of a quadratic function? A. If there are no x-intercepts and the y-intercept is positive, then the graph opens upwards. Thus the graph must open downwards, not upwards.

6 Version PREVIEW HW 01 hoffman 575) 6 keywords: CalC1a5s points Find the domain of the equation 1 x + x 4 = x. 1. Domain = {x : x 0, x 6}. Domain = {x : x > 0, x 4} 3. Domain = {x : x < 0, x 6} 4. Domain = {x : x 6, 4 } 5. Domain = {x : x 0, x 4} correct The domain of an equation consists of all values of x for which the expressions in the equation are well defined. Now, for the equation above, the right hand side is well defined for all non-negative x, while the left hand side is well-defined except at the zeros of the denominator. But x + x 4 = x + 6)x 4), so the zeros of the denominator of the left hand side occur at x = 6, 4, and Domain = {x : x 0, x 4}.