What are the x and y intercepts for #y = 1/2 (x-4)^2 +18#?

1 Answer
Mar 6, 2018

Answer:

There is no #x#-intercept. #y#-intercept is #26#.

Explanation:

To find #x#-intercept of any curve, just put #y=0#

and to #x#-intercept of any curve, just put #x=0#.

Hence #x#-intercept of #y=1/2(x-4)^2+18# is given by #1/2(x-4)^2+18=0# or #1/2(x-4)^2=-18#. But this is not possible asLHS cannot be negative. Hence, we do not have #x#-intercept.

For #y#-intercept of #y=1/2(x-4)^2+18#, put #x=0# and then #y=1/2*(-4)^2+18=26#. Hence #y#-intercept is #26#.

graph{y=1/2(x-4)^2+18 [-77, 83, -18.56, 61.44]}