Thread: How Do I Find the x and y intercepts??

Thanks in advance

To find the x intercept, solve for
To find the y intercept solve when x = 0

Note that there may not necessarily be an x or y intercept

Originally Posted by redsox25

Thanks in advance

Since that is the same as . There is no x intercept because ln(0) is not defined. To find the y intercept, use the fact that ln(1)= 0. f(1)= 2(1)ln(1)= 0. If x> 1, 2x ln(x)> 0 and if x< 1, 2xln(x)< 0 so that is the only y-intercept.

so when

it's

Therefore and









but when i graph it x=0 doesnt exist! why is this??

To find the y intercept, use the fact that ln(1)= 0. f(1)= 2(1)ln(1)= 0. If x> 1, 2x ln(x)> 0 and if x< 1, 2xln(x)< 0 so that is the only y-intercept.

While it is true that ln(x) > 0 and ln(x²) = 2ln(x), the question did ask for the intercept of ln(x²), not 2ln(x). There is a difference, for example has no y intercept but if you simplify it to y = x it does have one.

In which case we also need to consider the negative values of x because it is squared to become positive within the natural logarithm function.

but when i graph it x=0 doesnt exist! why is this??

This is because ln(0) is undefined. the point (0,0) becomes a point of discontinuity. When graphing, you draw a hollow circle around the point to indicate this.

Originally Posted by redsox25

so when

it's

Therefore and









but when i graph it x=0 doesnt exist! why is this??

The limiting value of your function as x --> 0 is 0 but it's not defined at x = 0 for reasons given earlier. The graph will have a 'hole' at (0, 0).