How do you graph #f( x ) = \frac { x + 3} { x ^ { 2} - 2x - 63}#?

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Answer 1 · 2018-01-23T23:29:17

asymptotes: #x = -7, x = 9#
#x-#intercept: #(3,0)#
#y-#intercept: #(0, - 1/21)#

firstly, factorise #x^2-2x-63#:

#7+( -9) = -2#

#7 *( -9) = -63#

#x^2-2x-63 = (x+7)(x-9)#

#(x+3)/(x^2-2x-63) = (x-3)/((x+7)(x-9))#

in this form, it is easier to find any asymptotes that there may be.

noting that #n/0=# undefined:

#(x-3)/(x+7)# is undefined
when #x=-7, x+7 = 0#
this means that the graph cannot go through the line #x=-7#.

#(x-3)/(x-9)# is undefined
when #x=9, x-9 = 0#
this means that the graph cannot go through the line #x= 9#.

to plot the #x-#intercept:

use #0/n = 0#

the numerator in the function is #x-3#.

#y = 0# when #x-3=0#, so #y=0# when #x=3#.

this means that the coordinates of the #x-#intercept are #(3,0)#.

to plot the #y-#intercept:

set #x# to #0#.

#(x-3)/((x+7)(x-9)) = (-3)/(7*-9) = -(3)/(63) = -1/21#

when #x = 0#, #y = - 1/21#

this means that the coordinates of the #y-#intercept are #(0, - 1/21)#

graph{(x-3)/((x+7)(x-9)) [-10, 10, -5, 5]}