Respuesta :
We have that [tex]y= \frac{x}{x^{2}-1 } [/tex], so we can factor the denominator to get:
[tex]y= \frac{x}{(x-1)(x+1)} [/tex]
Now, when x=0 [tex]y= \frac{0}{(0-1)(0+1)} [/tex], y=0; therefore the y-intercept is zero.
And, when y=0 [tex]0= \frac{x}{(x-1)(x+1)} = x=[0[(x-1)(x+1)]][/tex], x=0; therefore the x-intercept is also zero.
We can conclude that both the y-intercept and the x-intercept are equal zero.
[tex]y= \frac{x}{(x-1)(x+1)} [/tex]
Now, when x=0 [tex]y= \frac{0}{(0-1)(0+1)} [/tex], y=0; therefore the y-intercept is zero.
And, when y=0 [tex]0= \frac{x}{(x-1)(x+1)} = x=[0[(x-1)(x+1)]][/tex], x=0; therefore the x-intercept is also zero.
We can conclude that both the y-intercept and the x-intercept are equal zero.
Given the function, y = x/( x^2 – 1)
Add x and subtract x from (x^2 – 1) to make it a complete square
y = x/(x^2 + x – x – 1)
y = x/[x(x + 1) – 1(x + 1)]
y = x/[(x – 1)(x + 1)]
In order to find the y -intercept, we set x = 0
When x = 0
y = 0/[(0 – 1)(0 + 1)]
y = 0.
Therefore, the y-intercept is zero.
In order to find the x -intercept, we set y = 0
When y = 0
0 = x/[(x – 1)(x + 1)]
The only way the product of a division will be zero is if the numerator is zero
So, if x/[(x – 1)(x + 1)] = 0
x = 0
Therefore, the x-intercept is also zero.
Since both x- intercept and y- intercept equal zero,
Then, the given function has an x-intercept that is equal to its y-intercept