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=(0+2(0)-3)/(0+3)`

`y\text{ = }(-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=(x^2+3x-x-3)/(x+3)`
`0=(x(x+3)-1(x+3))/(x+3)`
`0=((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