Question

1. let ( g(x)=\frac{(x - 2)^3-x^3}{2x^2+3x - 1} ). which of the following statements about the graph of ( g ) is correct?

(a) the graph of ( g ) has a horizontal asymptote of ( y=\frac{3}{2} ).
(b) the graph of ( g ) has a horizontal asymptote of ( y=-1 ).
(c) the graph of ( g ) has a horizontal asymptote of ( y=-3 ).
(d) the graph of ( g ) does not have a horizontal asymptote.

1. let ( h(x)=\frac{x^3+x^2-20x}{x^2-8x + 16} ). which of the following statements about the graph of ( h ) is correct when ( x = 4 )

(a) the graph of ( h ) only has a zero when ( x = 4 ).
(b) the graph of ( h ) only has a hole when ( x = 4 ).
(c) the graph of ( h ) only has a vertical asymptote when ( x = 4 ).
(d) the graph of ( h ) has both a zero and a vertical asymptote when ( x = 4 ).

1. let ( m(x)=\frac{-4(x + 2)(x - 3)}{(x - 3)(x - 1)} ). then ( lim _{x

ightarrow3}m(x)=)
(a) (-10) (b) (-4) (c) (0) (d) ( infty )

Explanation:

Problem 15

Step1: Expand the numerator

$$(x - 2)^3-x^3=(x^3-6x^2 + 12x-8)-x^3=-6x^2+12x - 8$$

Step2: Analyze the degrees of the numerator and denominator

The degree of the numerator \(n=-6x^2+12x - 8\) is \(n = 2\), and the degree of the denominator \(d = 2x^2+3x - 1\) is \(d=2\)

Step3: Find the horizontal - asymptote formula

When \(n = d\), the horizontal asymptote \(y=\frac{a_n}{b_d}\), where \(a_n\) is the leading coefficient of the numerator and \(b_d\) is the leading coefficient of the denominator. Here \(a_n=-6\) and \(b_d = 2\)
$$y=\frac{-6}{2}=-3$$

Step1: Factor the numerator and denominator

• Numerator: \(x^3+x^2-20x=x(x^2 + x-20)=x(x + 5)(x-4)\)
• Denominator: \(x^2-8x + 16=(x - 4)^2\)

Step2: Simplify the function

\(h(x)=\frac{x(x + 5)(x-4)}{(x - 4)^2}=\frac{x(x + 5)}{x-4},x
eq4\)

Step3: Analyze the behavior at \(x = 4\)

Since the factor \((x - 4)\) cancels out in the simplification (but \(x
eq4\) in the original function), there is a hole at \(x = 4\)

Step1: Simplify the function

Since \(x
eq3\) (when taking the limit as \(x\to3\)), we can cancel out the non - zero factor \((x - 3)\) (for \(x
eq3\)) in the numerator and denominator.
\(m(x)=\frac{-4(x + 2)(x-3)}{(x - 3)(x-1)}=\frac{-4(x + 2)}{x-1},x
eq3\)

Step2: Substitute \(x = 3\) into the simplified function

\(\lim_{x\to3}m(x)=\frac{-4(3 + 2)}{3-1}=\frac{-4\times5}{2}=-10\)

Answer:

C. The graph of \(g\) has a horizontal asymptote of \(y =-3\)

Problem 16