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

1. ## How Do I Find the x and y intercepts??

$\displaystyle f(x)=xln(x^2)$

2. To find the x intercept, solve for $\displaystyle f(x) = 0$
To find the y intercept solve $\displaystyle y = xln(x^2)$ when x = 0

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

3. Originally Posted by redsox25
$\displaystyle f(x)=xln(x^2)$
Since $\displaystyle ln(x^2)= 2ln(x)$ that is the same as $\displaystyle f(x)= 2xln(x)$. 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.

4. so when $\displaystyle f(x)=0$

it's $\displaystyle 0=xln(x^2)$

Therefore $\displaystyle 0=x$ and $\displaystyle 0= \ln (x^2)$

$\displaystyle 0= \ln(x^2)$

$\displaystyle \Rightarrow e^0 = e^{\ln (x^2)}$

$\displaystyle \Rightarrow 1=x^2$

$\displaystyle \Rightarrow x=-1, 1$

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

5. 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 $\displaystyle y = \frac{x^2}{x}$ 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.

6. Originally Posted by redsox25
so when $\displaystyle f(x)=0$

it's $\displaystyle 0=xln(x^2)$

Therefore $\displaystyle 0=x$ and $\displaystyle 0= \ln (x^2)$

$\displaystyle 0= \ln(x^2)$

$\displaystyle \Rightarrow e^0 = e^{\ln (x^2)}$

$\displaystyle \Rightarrow 1=x^2$

$\displaystyle \Rightarrow x=-1, 1$

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).