Answered

Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

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]
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.