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

I need help with this problem, I need the hole, the horizontal asymptote , domain-example-1

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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

User Enkhbat
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