hello.
please graph these two
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
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
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
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