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Math  /  Algebra

QuestionThe function f(x)=1x+4−3f(x)=\frac{1}{x+4}-3 is a rational function. a. Use transformations of y=1xy=\frac{1}{x} or y=1x2y=\frac{1}{x^{2}} to sketch the graph. b. Find all xx-intercepts or state that the function has no xx-intercepts. c. Find the yy-intercept or state that the function does not have a yy-intercept. d. Find the equation(s) of all vertical asymptotes. e. Find the equation(s) of all horizontal asymptotes.

Studdy Solution

STEP 1

1. The function f(x)=1x+4−3 f(x) = \frac{1}{x+4} - 3 is a transformation of the basic rational function y=1x y = \frac{1}{x} .
2. Transformations include horizontal shifts, vertical shifts, and reflections.
3. The function can have intercepts and asymptotes based on its form.

STEP 2

1. Identify transformations of the basic function y=1x y = \frac{1}{x} .
2. Determine the x x -intercepts.
3. Determine the y y -intercept.
4. Identify vertical asymptotes.
5. Identify horizontal asymptotes.

STEP 3

Identify transformations of y=1x y = \frac{1}{x} to obtain f(x)=1x+4−3 f(x) = \frac{1}{x+4} - 3 :
- The term x+4 x+4 indicates a horizontal shift left by 4 units. - The term −3-3 indicates a vertical shift down by 3 units.
Thus, the graph of f(x) f(x) is the graph of y=1x y = \frac{1}{x} shifted left by 4 units and down by 3 units.

STEP 4

Determine the x x -intercepts by setting f(x)=0 f(x) = 0 :
1x+4−3=0 \frac{1}{x+4} - 3 = 0
Add 3 to both sides:
1x+4=3 \frac{1}{x+4} = 3
Multiply both sides by x+4 x+4 :
1=3(x+4) 1 = 3(x+4)
1=3x+12 1 = 3x + 12
Subtract 12 from both sides:
−11=3x -11 = 3x
Divide by 3:
x=−113 x = -\frac{11}{3}
Thus, the x x -intercept is (−113,0) \left(-\frac{11}{3}, 0\right) .

STEP 5

Determine the y y -intercept by evaluating f(0) f(0) :
f(0)=10+4−3 f(0) = \frac{1}{0+4} - 3
f(0)=14−3 f(0) = \frac{1}{4} - 3
f(0)=14−124 f(0) = \frac{1}{4} - \frac{12}{4}
f(0)=−114 f(0) = -\frac{11}{4}
Thus, the y y -intercept is (0,−114) \left(0, -\frac{11}{4}\right) .

STEP 6

Identify vertical asymptotes by finding values of x x that make the denominator zero:
The denominator x+4=0 x+4 = 0 when x=−4 x = -4 .
Thus, there is a vertical asymptote at x=−4 x = -4 .

STEP 7

Identify horizontal asymptotes by considering the behavior as x→±∞ x \to \pm \infty :
As x→±∞ x \to \pm \infty , 1x+4→0 \frac{1}{x+4} \to 0 .
Thus, f(x)=1x+4−3→−3 f(x) = \frac{1}{x+4} - 3 \to -3 .
Therefore, the horizontal asymptote is y=−3 y = -3 .
The graph transformations, intercepts, and asymptotes have been determined as follows: - Transformations: Left 4 units, down 3 units. - x x -intercept: (−113,0) \left(-\frac{11}{3}, 0\right) . - y y -intercept: (0,−114) \left(0, -\frac{11}{4}\right) . - Vertical asymptote: x=−4 x = -4 . - Horizontal asymptote: y=−3 y = -3 .

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