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Math 106 Name Graded Homework For #1-4: f(x) = x-5, g(x) = x - x - 6, h(x) = 4x + 3, and p(x) = 9x2 - 25 1. Find the x-intercept(s) for: 2. Find the y-intercept for: a) f(x) b) g(x) a) f(x) b) g(x) c) h(x) 3. Find: a) f(2) b) g(-3) c) (3) d) g(x - 2) e) (g)(2) f) (poh)(-2) g) (g+ p)(x) h) (f-g)(x) Dop)(x) 4. Find the domain of each function. Write your answer in interval notation. a) g(x) b) h(x) d) poh)(x) 5. Find the domain of each function. Write your answer in interval notation a) f(x) = (2x - 4) b) f(z) = DE DE 6. Solve each inequality. Write the solution in interval notation a) 6x - 5 5 4 b) |-3x - 11 >5 Solution Solution 7. Write the equation of the line that write your answer in slope-intercept form) a) is vertical and passes b) has a slope of 2 and c) passes through through (8,-7) x-intercept 9 (5.-1) and (4,4) d) is parallel to y = -x + 2 and and passes through (-2,-4) e) is perpendicular to 2x + y = 7 and passes through (7.-3) 8. Find the roots of the following functions: a) f(x) = 4x - 5 b) f(x) = 3x - 75x c)f(x) = x - 16x + 63r d)f(x) = 7x - 28 e) f(x) = -5x + 2 9. On your calculator, graph /(x) = -3r? +6. a) What is the range of f(x)? Write your answer in interval notation b) Does f(x) have an inverse7 How do you know? 10. Each function below has an inverse. Find f(x). a)/() = 7x-3 b) f(x) = (x - 1)' +8 11. Solve each equation for a) 432-1 = 47b) 16" . c) x3 = 716 d) xi = 27°) (17)* = 49 12. Calculate without a calculator a) b) 21:27 13 c)64 d) Ine e) logs logs 125 8) log, 162 + logo h) in (ve.es) fG) 13. Refer to the graph at the right Find: a) ft2) b) the domain the range d) For what value of x does f(x) attain a maximum? w e) What is the maximum value? $ 1) What x value(s) are the zeros of the function? 8) For what value(s) of x does f(x) =-1? 14. Graph: a) the parabola f(x) = -x - 3x + 4. Plot the roots, the maximum or minimum point, and the y-intercept b) f(x) = { wien - 35=
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I solve all questions, please read. Some text was not readable partially, I had to guess.docx and pdf files are the same

𝑓(𝑥) = 𝑥 − 5, 𝑔(𝑥) = 𝑥 2 − 𝑥 − 6, ℎ(𝑥) = √4𝑥 + 3, 𝑝(𝑥) = 9𝑥 2 − 25.
1. x-intercepts are points on graph where y=0.
a) 𝑓(𝑥) = 0, 𝑥 − 5 = 0, 𝑥 = 5. The only x-intercept is (𝟓, 𝟎).
b) 𝑔(𝑥) = 𝑥 2 − 𝑥 − 6 = 0, guess 𝑥1 = 3 and by Viet’s theorem 𝑥2 = −2.
The intercepts are (−𝟐, 𝟎), (𝟑, 𝟎).

2. y-intercepts are points on graph where x=0.
a) 𝑓(0) = −5, the y-intercept is (𝟎, −𝟓).
b) 𝑔(0) = −6, the y-intercept is (𝟎, −𝟔).
c) ℎ(0) = √3, the y-intercept is (𝟎, √𝟑).

3.
a) 𝑓(2) = 2 − 5 = −𝟑,
b) 𝑔(−3) = (−3)2 − (−3) − 6 = 𝟔,
c) ℎ(3) = √12 + 3 = √𝟏𝟓,
d) 𝑔(𝑥 − 2) = (𝑥 − 2)2 − (𝑥 − 2) − 6 = 𝑥 2 − 4𝑥 + 4 − 𝑥 + 2 − 6 = 𝒙𝟐 − 𝟓𝒙,
e) 𝑓(𝑔(2)) = 𝑓(22 − 2 − 6) = 𝑓(−4) = −4 − 5 = −𝟗,
f) 𝑝(ℎ(−2)) does not exist because ℎ is not defined at 𝑥 = −2 (4 ∙ (−2) + 3 = −5 < 0),
g) (𝑔 + 𝑝)(𝑥) = 𝑔(𝑥) + 𝑝(𝑥) = 𝑥 2 − 𝑥 − 6 + 9𝑥 2 − 25 = 𝟏𝟎𝒙𝟐 − 𝒙 − 𝟑𝟏,
h) (𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥) = 𝑥 − 5 − 𝑥 2 + 𝑥 + 6 = −𝒙𝟐 + 𝟐𝒙 + 𝟏,
i) (𝑓 ∙ 𝑝)(𝑥) = 𝑓(𝑥)𝑝(𝑥) = (𝑥 − 5)(9𝑥 2 − 25) = 𝟗𝒙𝟑 − 𝟒𝟓𝒙𝟐 − 𝟐𝟓𝒙 + 𝟏𝟐𝟓.

4.
a) (−∞, ∞),
3

3

b) 4𝑥 + 3 ≥ 0, 𝑥 ≥ − 4 , [− 4 , ∞),
𝑝

𝑝(𝑥)

c) 𝑓 (𝑥) = 𝑓(𝑥), the only restriction is 𝑓(𝑥) ≠ 0, 𝑥 ≠ 5, 𝑥 ∈ (−∞, 𝟓) ∪ (𝟓, ∞).
3

𝟑

d) 𝑝 works with any 𝑥 so the only restriction is 4𝑥 + 3 ≥ 0, 𝑥 ≥ − 4 , [− 𝟒 , ∞).

5.
1
a) 𝑓(𝑥) = (2𝑥 − 4)−3 = (2𝑥−4)3 , the only restriction is 2𝑥 − 4 ≠ 0, 𝑥 ≠ 2, 𝑥 ∈ (−∞, 𝟐) ∪ (𝟐, ∞),
b) 𝑓(𝑥) =

√𝑥−3
.
𝑥−8

It must be 𝑥 − 3 ≥ 0, 𝑥 − 8 ≠ 0, i.e. 𝑥 ≥ 3, 𝑥 ≠ 8, 𝑥 ∈ [𝟑, 𝟖) ∪ (𝟖, ∞),

6.
a) |6𝑥 − 5| ≤ 4.
5
1
For 6𝑥 − 5 ≤ 0, 𝑥 ≤ 6 we have −(6𝑥 − 5) ≤ 4, −6𝑥 ≤ −1, 𝑥 �...


Anonymous
Very useful material for studying!

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