Question

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

What is the domain and range, x-and y-intercepts, horizontal asymptotes, and vertical asymptotes?

Answer 1

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

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

Answer 2

i. Domain and Range

The given function is

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

The domain of this function is,

`x^2-3x-4\\e 0`

`(x-4)(x+1)\\e 0`

`x\\e4,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;

`(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)=(0^2+0-2)/(0^2-3(0)-4)`

`f(0)=(1)/(2)`

The y-intercept is

`(0,(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=(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`