# Math Help - Finding intercepts

1. ## Finding intercepts

ok so i need to find the intercepts, includes x and y, of this equation y=(x-1) times the square root of (x^2+1)
so far when i try doing it i don't get the points which are shown in the calculator

2. Originally Posted by la_lo626
ok so i need to find the intercepts, includes x and y, of this equation y=(x-1) times the square root of (x^2+1)
so far when i try doing it i don't get the points which are shown in the calculator
To find the y-intercept, set x=0.

$y=(0-1)\sqrt{0^2+1} = -1$

The y-intercept is at point (0,-1).

To find the x-intercept, set y=0.

$0=(x-1)\sqrt{x^2+1}$

Since $\displaystyle\sqrt{x^2+1}$ can't equal 0, we must have x = 1.

The x-intercept is at point (1,0).

3. Originally Posted by undefined
To find the y-intercept, set x=0.

$y=(0-1)\sqrt{0^2+1} = -1$

The y-intercept is at point (0,-1).

To find the x-intercept, set y=0.

$0=(x-1)\sqrt{x^2+1}$

Since $\displaystyle\sqrt{x^2+1}$ can't equal 0, we must have x = 1.

The x-intercept is at point (1,0).

i am confused in the x-intercept part. the answer is cosrrect but i was wondering how did you get to that point i'm just not sure is there a step in which you're sppose to factor out the equation?

4. Originally Posted by la_lo626
i am confused in the x-intercept part. the answer is cosrrect but i was wondering how did you get to that point i'm just not sure is there a step in which you're sppose to factor out the equation?
Suppose "a" and "b" are real numbers and ab=0. Then either a=0 or b=0 (or both). Can you see why $\displaystyle\sqrt{x^2+1}$ can't possibly be 0?

So $x-1 = 0 \implies x=1$.

5. thanks "undefined" i see now sort of it's a little more clear