# How do you compare and contrast the graphs of #y=x-2# and #y=(x^2+5x-14)/(x+7)#?

##### 1 Answer

Jan 29, 2018

#### Explanation:

#"consider "y=(x^2+5x-14)/(x+7)#

#"factorising the numerator gives"#

#y=((x+7)(x-2))/(x+7)#

#"cancelling the factor "(x+7)" from numerator/denominator"#

#y=(cancel((x+7))(x-2))/cancel((x+7))=x-2#

#"the removal of "(x+7)" from the numerator/denominator"#

#"indicates there is a hole at "x=-7#

#rArry=(x^2+5x-14)/(x+7)" reduces to "y=x-2#

#"the graphs of "y=(x^2+5x-14)/(x+7)" and "y=x-2#

#"are the same graph "#

#"the only difference being that "y=(x^2+5x-14)/(x+7)#

#"has a hole at "x=-7" whereas "y=x-2" does not"#