Question

I need help with this problem, I need the hole, the horizontal asymptote , domain and x intercept

Answer 1

The function

$f(x)=\frac{x^2+2x\text{ -3}}{x+3}$

let f(x) = y

$y\text{ =}\frac{x^2+2x\text{ - 3}}{x+3}$

substitute x = 0

$y=\frac{0+2(0)-3}{0+3}$

$y\text{ = }\frac{-3}{3}$$y\text{ = -1}$

(x = 0 and y = -1)

Next, put y = 0 in the function and then solve for x

$0=\frac{x^2+2x\text{ -3}}{x+3}$

$0=\frac{x^2+3x-x-3}{x+3}$$0=\frac{x(x+3)-1(x+3)}{x+3}$$0=\frac{(x+3)(x-1)}{x+3}$$0=x-1$$x=1$

(x = 1 and y =0)

The domain is (0 and 1)

The x -intercept is (1, 0)

The hole is

x + 3 =0

x= -3 is the hole

To find the horizontal asymptote;

since the degree of the numerator is greater than the denominator, then it has no horiizontal asymptote

horizontal asymptote = none