of multiplicity two. The sign of the polynomial is shown in the table below
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1 161 Precalculus 1 Review 5 Problem 1 Graph the polynomial function P( ) ( ) ( 1). Solution The polynomial is of degree 4 and therefore it is positive to the left of its smallest real root and to the right of its largest real root. The roots of the polynomial are,0, and 1. The sign of the polynomial changes at and at 1, because they are simple roots, and does not change at0 because 0 is a root of multiplicity two. The sign of the polynomial is shown in the table below Interval (, ) (,0) (0,1) (1, ) Sign The graph of the polynomial is shown below.
2 Problem Consider the polynomial (a) Find the -intercepts. (b) Find the critical points and the range. (c) Graph the polynomial 4 P( ) 1. Solution (a) to find the -intercepts we have to find the real solutions of the 1 0. This equation is of quadratic type and applying the equation 4 quadratic formula we get ( ) ( ) 4 1 ( 1) 1 a Sign minus does not provide any real solutions and taking sign plus we obtain (b) To find the range and the position of critical points we have to solve the equation that 4 y 1for. Applying again the quadratic formula we see 9 4( y 1) y. The epression under square root cannot be negative whence Therefore y and the range of Pis[ 1 / 4, ). 4
3 There are critical points corresponding to the value y 1 4. Respectively and 1. there is one more critical point at(0, 1). The list of critical points is. Because polynomial Pis an even function (, 1 4),(0, 1),(, 1 4) (c) The graph of Pis shown below.
4 Problem Consider the linear fraction R ( ) 4. (a) Find the and y -intercepts (b) Find equations of the horizontal and the vertical asymptote (c) Graph the function together with its horizontal and vertical asymptotes (d) Find an equation of the inverse function and graph it together with its horizontal and vertical asymptotes Solution (a) to find the -intercept we solve the equation y 0 equivalent to 0 whence and the intercept is which is (,0). To find the y -intercept we just notice that if 0 then y 4and the y -intercept is(0, 4). (b) Looking at the ratio of leading terms horizontal asymptote is y. we see that an equation of the The function is undefined if 4whence an equation of the vertical asymptote is 4.
5 (c) The graph is shown below (d) solving the equation y 4 whence (y ) 4y and for we get y 4y 1 4 R ( ) 4y y. The inverse function is The and y -intercepts are, respectively, ( 4,0) and(0, horizontal asymptote is 4 graph is shown below. ). The y, the vertical asymptote is. The
6
7 Problem 4 Consider the rational function (a) Find the and the y -intercepts, if any R ( ) (b) Find an equation of the horizontal and the slant asymptote (c) Find the critical points, if any, and the range of the function (d) Graph the function together with its horizontal and slant asymptotes 1 1. Solution (a) the equation no -intercepts. The y -intercept i (0,1). 1 0has no real solutions whence there are (b) The vertical asymptote is clearly 1. To find an equation of the slant asymptote we will divide 1by 1 using e.g. the synthetic division The quotient is whence an equation of the slant asymptote is y (c)to find the range and the critical points (if they eist) we have to solve the 1 equation y 1 quadratic equation (1 y) (1 y ) 0. The quadratic formula provides for. This equation is equivalent to the following y y y y y y 1 ( 1) 4( 1) 1 ( 1)( ) The epression is defined if and only if ( y 1)( y ) 0which happens when either y 1or y. The range of Ris therefore
8 (, ] [1, ) To find the critical points we plug in into the epression for the values y 1and y getting 0 and are located at (, )and at(0,1). (d) The graph of R and its asymptotes is shown below, respectively. The critical points
9 Problem 5 Consider the function R 4 1 (a) Find the -intercepts (b) Find the vertical asymptotes (c) Find the horizontal or slant asymptote, if any (d) Find the sign of the function (e) Graph the function and its asymptotes Solution (a) the -intercepts are (,0)and(,0) (b) the vertical asymptotes are 1and 1 (c) the degree of the numerator equals the degree of the denominator; the ratio of the leading terms is 1; therefore there is the horizontal asymptote with the equation y 1 (d) we see from (c) that the sign of Ris positive far right and far left. Because R ( ) ( )( ) ( 1)( 1) every root of the numerator and of the denominator is simple and the sign of the function changes at points, 1,1 and. Interval (, ) (, 1) ( 1,1) (1,) (, ) Sign of R
10 (f) The graph is shown below
( ) = 1 x. g( x) = x3 +2
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