i. Domain and Range
The given function is
[tex]f(x)=\frac{x^2+x-2}{x^2-3x-4}[/tex]
The domain of this function is,
[tex]x^2-3x-4\ne 0[/tex]
[tex](x-4)(x+1)\ne 0[/tex]
[tex]x\ne4,xne -1[/tex]
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 [tex]f(x)=0[/tex]
This implies that;
[tex]\frac{x^2+x-2}{x^2-3x-4}=0[/tex]
This will give us
[tex]x^2+x-2=0[/tex]
[tex]\Rightarrow x^2+x-2=0[/tex]
[tex]\Rightarrow x^2+2x--x-2=0[/tex]
[tex]\Rightarrow x(x+2)-1(x+2)=0[/tex]
[tex]\Rightarrow (x+2)(x-1)=0[/tex]
[tex]\Rightarrow (x+2)=0,(x-1)=0[/tex]
[tex]\Rightarrow x=-2,x=1[/tex]
The x-intercepts are [tex](-2,0),(1,0)[/tex]
For y-intercept, we put
[tex]x=0[/tex] to obtain;
[tex]f(0)=\frac{0^2+0-2}{0^2-3(0)-4}[/tex]
[tex]f(0)=\frac{1}{2}[/tex]
The y-intercept is
[tex](0,\frac{1}{2})[/tex]
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 [tex]y=\frac{1}{1}=1[/tex]
iv. Vertical asymptote
To find the vertical asymptote, we equate the denominator to zero to get;
[tex]x^2-3x-4=0[/tex]
This implies that;
[tex]x^2+x-4x-4=0[/tex]
Split the middle term
[tex]x(x+1)-4(x+1)=0[/tex]
Factor
[tex](x+1)(x-4)=0[/tex]
Factor further
[tex](x+1)=0,(x-4)=0[/tex]
[tex]x=-1,x=4[/tex]
The vertical asymptotes are [tex]x=-1,x=4[/tex]
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
