Finding intercepts

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July 7th 2010, 08:42 AM
la_lo626

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
July 7th 2010, 08:47 AM
undefined

Quote:

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).
July 7th 2010, 08:57 AM
la_lo626

Quote:

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?
July 7th 2010, 09:05 AM
undefined

Quote:

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 $x-1 = 0 \implies x=1$ .
July 7th 2010, 09:12 AM
la_lo626

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