Set y = 0?

**tsmith** · October 2nd 2009, 06:29 AM

The question first tells me to graph the function to determine any x-intercepts.
y = 20(2/X+1 - 3/x)
When I put it in my calculator, I got x-intercept = (1,0)
Now it tells me to set y = 0 to confirm this result.
So,
0 = 20(2/X+1 - 3/x)
I started working in the brackets:
(2x)/(x)(x + 1) - (3x+1)/(x)(x+1)
(-x+1) / (x)(x+1)
Is this right?
If so, what do I do next? What do I do with the 20?

**Defunkt** · October 2nd 2009, 08:08 AM

Originally Posted by tsmith

The question first tells me to graph the function to determine any x-intercepts.
y = 20(2/X+1 - 3/x)
When I put it in my calculator, I got x-intercept = (1,0)
Now it tells me to set y = 0 to confirm this result.
So,
0 = 20(2/X+1 - 3/x)
I started working in the brackets:
(2x)/(x)(x + 1) - (3x+1)/(x)(x+1)
(-x+1) / (x)(x+1)
Is this right?
If so, what do I do next? What do I do with the 20?

$\frac{2x}{x(x+1)} - \frac{3x+1}{x(x+1)} \neq \frac{2}{x+1} - \frac{3}{x} = \frac{2x}{x(x+1)} - \frac{3(x+1)}{x(x+1)}$ $= \frac{2x - 3(x+1)}{x(x+1)} = \frac{-x-3}{x(x+1)}$

Can you solve $\frac{-3-x}{x(x+1)} = 0$ ?

**tsmith** · October 2nd 2009, 09:03 AM

-3/(x+1)?

**e^(i*pi)** · October 2nd 2009, 10:47 AM

Originally Posted by tsmith

-3/(x+1)?

$x = -3$

Take tsmith's final working:

$<br /> \frac{-3-x}{x(x+1)} = 0<br />$

Multiply both sides by $x(x+1)$

$-3-x = 0$

Can you see $x = -3$ now?

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