# please graph these 2 graphs

• Mar 5th 2007, 08:45 PM
girl1985
hello.

f(x)=(x^2-1)(1-x/2), also if you could find the real zeros of this polynomial

f(x)=x^2+3x+2/x-1

thanks
• Mar 5th 2007, 10:32 PM
earboth
Quote:

Originally Posted by girl1985
hello.

f(x)=(x^2-1)(1-x/2), also if you could find the real zeros of this polynomial
...

Hi,

you can factor the term of the function completely:

f(x) = (x² - 1)(1 - x/2) = (x - 1)(x + 1)(x - x/2)

A product equals zero if one of the factors is zero. Therefore solve for x;

x + 1 = 0 or x - 1 = 0 or 1 - x/2 = 0. You'llget the zeros: -1, 1, 2

I've attached the graph of the function.

EB
• Mar 5th 2007, 10:49 PM
earboth
Quote:

Originally Posted by girl1985
hello.

...
f(x)=x^2+3x+2/x-1

Hi,

I'm not quite certain how to read this problem. I assume that you mean:

f(x)=x^2+3x+2/(x-1)

The graph has a vertical asymptote at x = 1 because the function is not defined at x = 1.

To calculate the zeros you have to solve an equation of 3rd degree:

x³ + 2x² - 3x + 2 = 0. The solution is possible if you use Cardano's formula. It's a lot of work so I let my computer do it.

You get only one zero at x -3.15275...

I've attached a diagram of the graph. The asymptote is painted red.

EB
• Mar 5th 2007, 10:56 PM
earboth
Quote:

Originally Posted by girl1985
hello.

f(x)=x^2+3x+2/x-1

thanks

Hi,

I'm not quite certain how to read this problem. I assume that you mean:

f(x)=x^2+3x+2/x-1

The graph has a vertical asymptote at x = 0 because the function is not defined at x = 0.

To calculate the zeros you have to solve an equation of 3rd degree:

x³ + 3x² - x + 2 = 0. The solution is possible if you use Cardano's formula. It's a lot of work so I let my computer do it.

You get only one zero at x = -3.4566...

I've attached a diagram of the graph. The asymptote is painted red.

EB