How do you identify all asymptotes or holes and intercepts for #f(x)=(2x^2+3)/(x^2+6x+8)#?

1 Answer
Dec 14, 2017

Answer:

#"see explanation"#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

#" solve "x^2+6x+8=0rArr(x+2)(x+4)=0#

#rArrx=-4" and "x=-2" are the asymptotes"#

#"horizontal asymptotes occur as"#

#lim_(xto+-oo),f(x)toc" ( a constant)"#

#"divide terms on numerator/denominator by the highest"#
#"power of x that is "x^2#

#f(x)=((2x^2)/x^2+3/x^2)/(x^2/x^2+(6x)/x^2+8/x^2)=(2+3/x^2)/(1+6/x+8/x^2)#

#"as "xto+-oo,f(x)to(2+0)/(1+0+0)#

#rArry=2" is the asymptote"#

#"holes occur if a common factor is eliminated from"#
#"the numerator/denominator. This is not the case here"#
#"hence there are no holes"#

#color(blue)"Intercepts"#

#• " let x = 0 for y-intercept"#

#• " let y = 0 for x-intercepts"#

#x=0toy=3/8larrcolor(red)"y-intercept"#

#y=0to2x^2+3=0rArrx^2=-3/2#

#"this has no real solutions hence no x-intercepts"#
graph{(2x^2+3)/(x^2+6x+8) [-10, 10, -5, 5]}