How do you find intercepts, extrema, points of inflections, asymptotes and graph #f(x)=(2x^2-5x+5)/(x-2)#?

Question

William Abdalla · Accepted Answer

Two asymptotes; two #x#-intercepts; one #y#-intercept; and two stationary points.

First of all do polynomial division to get your function in the nice form of: #f(x) = 2x + 1 + 3/(x-2)#. From this it is fairly clear where the asymptotes are, as #x# cannot equal #2# as #f(x)# would then be undefined. Similarly #f(x)# cannot equal #2x + 1#, as this would mean that #3/(x-2)# equals #0#, which it cannot as it implies #3=0#. Then differentiate this function, applying the chain rule and product rules where necessary. The derivative of #2x + 1# is of course #2#. The derivative of #3/(x-2)# is done by letting #y=3(x-2)^-1#. First apply the chain rule: say #t = x-2# and #w=t^-1#. #:. (dt)/(dx) = 1# and #(dw)/(dt) = -t^-2#. #:. (dw)/(dx) = -t^-2 = -(x-2)^-2# The product rule isn't really necessary as it's pretty clear you will end up with: #(dy)/(dx) = -3/(x-2)^2#. #:. f'(x) = 2 -3/(x-2)^2#. Letting #f'(x) = 0# next. #:. 3 = 2(x-2)^2# Simplify it down to arrive at #2x^2-8x+8 = 0#, which by the quadratic formula leaves us with: #x=(4-sqrt(6))/2# or #x=(sqrt(6)+4)/2#. Solving for #f((4-sqrt(6))/2)# and #f((sqrt(6)+4)/2)# yields #5-sqrt(6)# and #2sqrt(6) + 5#, which are maximum and minimum respectively. Finally calculate your intercepts, first by letting #f(x)=0# #:. -3 = (2x+1)(x-2)# #:. 2x^2-3x+1 =0# #:. x = 1/2# or #x=1#. Finally calculate #f(0)# which is simply #1-3/2#, therefore the #y#-intercept is where #y=-1/2#.

Emma Aldridge · Answer

undefined To find intercepts, set $ f(x) $ equal to zero and solve for $ x $. For extrema, find the critical points by setting the derivative of $ f(x) $ equal to zero and solve for $ x $. Points of inflection occur where the second derivative of $ f(x) $ changes sign or is zero. To find them, find the second derivative and solve for $ x $. To find asymptotes, determine the vertical asymptotes by finding where the denominator of $ f(x) $ is zero (except where the numerator is also zero, which would be holes in the graph). Horizontal asymptotes occur when the degree of the numerator is less than or equal to the degree of the denominator. To graph the function, use the intercepts, extrema, points of inflection, asymptotes, and any other relevant information.