2 6 Algebra question Please answer informat shown Given kr

Given k(r) = 2r^2 + 5r - 3/2r^2 - 9r + 10, answer the following questions: Round all misters to 3 decimal places as needed. If the answer doesn\'t exist, type DXE. a) The domain of k written in interval notation is: b) The k(r)-intercept is: c) The r-intercept(s) are: d) The vertical asymptotes of k are at r = e) k has a hole at the ordered pair: Write the hole as an ordered pair, and use the rounded input to find the rounded output f) The horizontal asymptote of k is:

Solution

(a) We have 2r² -9r+10 = 2r² -4r-5r+10= 2r(r-2) -5(r-2) = (r-2)(2r-5). Thus, the denominator is 0 when r=2 or r=5/2. Since division by 0 is not defined,the domain of k( r) is {r R:r2and r5/2} In interval notation, the domain of k( r) is (-, 2)U(5/2,).

(b) The k (r )-intercept is where r = 0. On substituting r = 0 in k ( r) = (2r²+5r-3)/(2r² -9r+10), we have the k ( r) - intercept = -3/10.

(c ) The r-intercept is where k(r) = 0.Then 2r²+5r-3= 0 or, 2r²+6r-r-3=0 or, 2r(r+3)-1(r+3) = (r+3)(2r-1) which equals 0 when r = -3 or, r = ½. Thus, the r-intercepts are -3 and ½.

(d) The vertical asymptotes can be determined by finding the roots of the denominator, which are 2 and 5/2. Thus, the vertical asymptotes are r = 2 and r = 5/2.

(e) We have k (r ) =(2r²+5r-3)/(2r² -9r+10)= (r+3)(2r-1)/(r-2)(2r-5).Since there is no factor common between the numerator and the denominator, there is no removable discontinuity. Thus, there is no hole.

(f ) The degrees of the numerator and the denominator are same and the coefficient of each of the numerator and the denominator is 2. Hence the horizontal asymptote is k( r) = 2/2 or, k( r) = 1.