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Let [tex]\( f \)[/tex] be the function whose graph is obtained by translating the graph of [tex]\( y=\frac{1}{x} \)[/tex] to the left 4 units and up 2 units.

a. Write an equation for [tex]\( f(x) \)[/tex] as a quotient of two polynomials.

b. Determine the zero(s) of [tex]\( f \)[/tex].

c. Identify the asymptotes of the graph of [tex]\( f(x) \)[/tex].

Sagot :

GinnyAnswer
Let's work through the given problem step-by-step:

### Part (a)
To find the equation of the function [tex]\( f(x) \)[/tex] that results from translating the graph of [tex]\( y = \frac{1}{x} \)[/tex]:

1. Translation to the left by 4 units:
- To translate the graph to the left 4 units, we replace [tex]\( x \)[/tex] with [tex]\( x + 4 \)[/tex]. This modifies the function to [tex]\( y = \frac{1}{x + 4} \)[/tex].

2. Translation up by 2 units:
- To translate the graph up by 2 units, we add 2 to the function. This modifies the function to [tex]\( y = \frac{1}{x + 4} + 2 \)[/tex].

Therefore, the equation for [tex]\( f(x) \)[/tex] is:
[tex]\[ f(x) = \frac{1}{x + 4} + 2 \][/tex]

### Part (b)
To determine the zeros of [tex]\( f(x) \)[/tex]:

1. Set the function equal to zero:
[tex]\[ f(x) = 0 \][/tex]
[tex]\[ \frac{1}{x + 4} + 2 = 0 \][/tex]

2. Solve for [tex]\( x \)[/tex]:
- Subtract 2 from both sides:
[tex]\[ \frac{1}{x + 4} = -2 \][/tex]
- Take the reciprocal of both sides:
[tex]\[ x + 4 = -\frac{1}{2} \][/tex]
- Solve for [tex]\( x \)[/tex]:
[tex]\[ x = -\frac{1}{2} - 4 \][/tex]
[tex]\[ x = -\frac{9}{2} \][/tex]

So, the zero of [tex]\( f(x) \)[/tex] is:
[tex]\[ x = -\frac{9}{2} \][/tex]

### Part (c)
To identify the asymptotes of the graph of [tex]\( f(x) \)[/tex]:

1. Vertical Asymptote:
- The vertical asymptote occurs where the function is undefined, which is when the denominator of the fraction is zero:
[tex]\[ x + 4 = 0 \][/tex]
[tex]\[ x = -4 \][/tex]
So, the vertical asymptote is:
[tex]\[ x = -4 \][/tex]

2. Horizontal Asymptote:
- The horizontal asymptote is determined by looking at the behavior of the function as [tex]\( x \)[/tex] approaches infinity or negative infinity. For large positive or negative [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{x + 4} \rightarrow 0 \][/tex]
Thus, the function [tex]\( f(x) \rightarrow 2 \)[/tex] as [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex].
So, the horizontal asymptote is:
[tex]\[ y = 2 \][/tex]

### Summary:

a. The equation for [tex]\( f(x) \)[/tex] is:
[tex]\[ f(x) = \frac{1}{x + 4} + 2 \][/tex]

b. The zero of [tex]\( f(x) \)[/tex] is:
[tex]\[ x = -\frac{9}{2} \][/tex]

c. The asymptotes are:
- Vertical asymptote: [tex]\( x = -4 \)[/tex]
- Horizontal asymptote: [tex]\( y = 2 \)[/tex]
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