9514 1404 393
Answer:
Step-by-step explanation:
A lot of math is about matching patterns.
For example, ...
g(x) = f(x -h) +k
means g(x) is the function f(x) translated right by h units and up by k units. This will be true for any expression of f(x).
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In this problem, f(x) = √x. We want to translate it left 6 units (h=-6)*, and up 4 units (k=4).
The notation above means that we will replace x with (x-h) = x+6. and we will add k = 4 to the result.
f(x) = √x
g(x) = f(x+6) +4
g(x) = √(x+6) +4 . . . . . . matches choice D
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* Left is the opposite of right, so left 6 units is the opposite of right 6 units. h=6 for <em>right 6 units</em>, so h=-6 for <em>left 6 units</em>. Then x-h = x-(-6) = x+6.
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<em>Comment on the graph</em>
I find it useful to see a picture with these things. In the attached graphing calculator output, the blue curve is left 6 and up 4 from the red curve. The blue curve is g(x); the red one is f(x).
E+(e-24)=126
2e-24=126
2e=150
e=75
So (s)he recieved 75 and 75-24=51 emails on those days.
<h3>
Short Answer: Yes, the horizontal shift is represented by the vertical asymptote</h3>
A bit of further explanation:
The parent function is y = 1/x which is a hyperbola that has a vertical asymptote overlapping the y axis perfectly. Its vertical asymptote is x = 0 as we cannot divide by zero. If x = 0 then 1/0 is undefined.
Shifting the function h units to the right (h is some positive number), then we end up with 1/(x-h) and we see that x = h leads to the denominator being zero. So the vertical asymptote is x = h
For example, if we shifted the parent function 2 units to the right then we have 1/x turn into 1/(x-2). The vertical asymptote goes from x = 0 to x = 2. This shows how the vertical asymptote is very closely related to the horizontal shifting.
Step-by-step explanation:
60 divided by 7 = approx. 8-9 min for each task.
7 tasks per hour
Answer:
See Explanation.
General Formulas and Concepts:
<u>Pre-Algebra</u>
<u>Algebra II</u>
- Log/Ln Property:
<u>Calculus</u>
Derivatives
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Chain Rule: ![\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28g%28x%29%29%5D%20%3Df%27%28g%28x%29%29%20%5Ccdot%20g%27%28x%29)
Derivative of Ln: ![\frac{d}{dx} [ln(u)] = \frac{u'}{u}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Bln%28u%29%5D%20%3D%20%5Cfrac%7Bu%27%7D%7Bu%7D)
Step-by-step explanation:
<u>Step 1: Define</u>
<u>Step 2: Differentiate</u>
- Rewrite:
- Rewrite [Ln Properties]:
- Differentiate [Ln/Chain Rule/Basic Power Rule]:
- Simplify:
- Rewrite:
- Combine:
- Reciprocate:
- Distribute:
