How do you find the zeros for the function #f(x)=(x^2-x-12)/(x-2)#?

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Answer 1 · 2017-04-04T06:12:56

#f(x)# has zeros at #-3# and #4#

#f(x) =(x^2-x-12)/(x-2)#

#= ((x+3)(x-4))/(x-2)#

#:. f(x) = 0 -> ((x+3)(x-4))/(x-2) =0#

Hence the zeros of #f(x)# occur when #(x+3)(x-4)=0#
Note: #f(x)# is undefined at #x=2#

#:.f(x) = 0# at #x=-3# and #x=4#

This can be seen from the grapg of #f(x)# below:

graph{((x+3)(x-4))/(x-2) [-18.01, 18.04, -9.01, 8.99]}

Answer 2 · 2017-04-04T11:02:53

X= 4 and x= -3

You can find the zeros only when the nominator is equal to 0

So it's not necessary to care about the denominator in here

#x^2-x-12=0#

You can solve by factorisation, quadratic formula, or completing square

# (x-4)(x+3) = 0 #

And you can solve it

You get x= 4 and x= -3