What are the #x# and #y# intercepts of the curve #3x^2-x+8=0#?

1 Answer
Feb 21, 2017

Answer:

There is no #x#-intercept and #y#-intercept is at #(0,6)#.

Explanation:

For #x#-intercept put #y=f(x)=0# i.e. #3x^2-x+6=0#

but as discriminant is #b^2-4ac=(-1)^2-4xx3xx6=-71#, we do not have a real solution and hence there is no #x#-intercept. Also observe that #3x^2-x+6=3(x-1/6)^2+71/12# and hence for all values of #x#, #3x^2-x+6>0# and hence no #x#-intercept.

For #y#-intercept, we should put #x=0# and then #y=f(0)=6# and #y#-intercept is at #(0,6)#
graph{3x^2-x+6 [-20, 20, -2, 18]}