Question

Use algebraic methods to prove that the given function has an x-intercept that is equal to its y-intercept. In your final answer, include all of your calculations.
y = x/x^2 - 1

Answer 1

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

Answer 2

We have that $y= \frac{x}{x^{2}-1 }$, so we can factor the denominator to get:
$y= \frac{x}{(x-1)(x+1)}$
Now, when x=0 $y= \frac{0}{(0-1)(0+1)}$, y=0; therefore the y-intercept is zero.
And, when y=0 $0= \frac{x}{(x-1)(x+1)} = x=[0[(x-1)(x+1)]]$, x=0; therefore the x-intercept is also zero.

We can conclude that both the y-intercept and the x-intercept are equal zero.