Question

F(x) = x^2+x-2/x^2-3x-4

Answer 1

i. Domain and Range

The given function is

$f(x)=\frac{x^2+x-2}{x^2-3x-4}$

The domain of this function is,

$x^2-3x-4\ne 0$

$(x-4)(x+1)\ne 0$

$x\ne4,xne -1$

The range refers to the y-values for which x is defined. x  is defined for all values of y.

The range is all real numbers. See graph

ii. x-and-y-intercept

For x- intercept intercept we put $f(x)=0$

This implies that;

$\frac{x^2+x-2}{x^2-3x-4}=0$

This will give us

$x^2+x-2=0$

$\Rightarrow x^2+x-2=0$

$\Rightarrow x^2+2x--x-2=0$

$\Rightarrow x(x+2)-1(x+2)=0$

$\Rightarrow (x+2)(x-1)=0$

$\Rightarrow (x+2)=0,(x-1)=0$

$\Rightarrow x=-2,x=1$

The x-intercepts are $(-2,0),(1,0)$

For y-intercept, we put

$x=0$ to obtain;

$f(0)=\frac{0^2+0-2}{0^2-3(0)-4}$

$f(0)=\frac{1}{2}$

The y-intercept is

$(0,\frac{1}{2})$

iii. Horizontal asyptote

Since degree of the numerator and the denominator are the same, there is a horizontal asymptote

To find the horizontal asymptote.

We divide the leading coefficient of the numerator by the leading  coefficient of the denominator.

The horizontal asymptote is $y=\frac{1}{1}=1$

iv. Vertical asymptote

To find the vertical asymptote, we equate the denominator to zero to get;

$x^2-3x-4=0$

This implies that;

$x^2+x-4x-4=0$

Split the middle term

$x(x+1)-4(x+1)=0$

Factor

$(x+1)(x-4)=0$

Factor further

$(x+1)=0,(x-4)=0$

$x=-1,x=4$

The vertical asymptotes are $x=-1,x=4$

Answer 2

Answer:

D : {x ∈ R : x≠-1 and x≠4}

R : {all real number}

The x intercepts are 2,-1    

The y intercept is 1/2

The horizontal asymptotes is y=1

x=-1  x=4

Step-by-step explanation:

f(x) = x^2+x-2/x^2-3x-4

We need to factor

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

       ----------------

     ( x-4) (x+1)

The denominator goes to zero at 4 and -1

The domain is the possible values for x.  X cannot make the denominator go to zero so it cannot be -1 or 4

D : {x ∈ R : x≠-1 and x≠4}

The range is the possible values for y.  It can take on any real numbers

R : {all real number}

The x intercepts is where it crosses the x axis.  This is where the numerator goes to zero

The x intercepts are 2,-1

The y intercepts are where it crosses the x axis.  Let x =0 and solve for y

y = (0+2) (0-1)               -2

       ---------------- = ------------------ = 1/2

     ( 0-4) (0+1)            -4

The y intercept is 1/2

The horizontal asymptotes is 1

Since the polynomial is the same degree in the numerator and the denominator,divide the coefficients

1/1 = 1

y=1

Vertical asymptoptes are where the denominator is zero

x=-1  x=4