Write an equation that has these characteristics: period =2π, Maximum = 12; midline: y= 11; y-intercept: (0, 11)
1 answer:
Answer:
Step-by-step explanation:
The sine or cosine function and their reciprocals has a period of 2π.
The maximum should be 12 and midline equal y=11.
This means the must have an amplitude of
a=12-11=1
If the y-intercept is (0,11) the y-intercept was translated up 11 units from the origin so it should be a sine graph.
The function is of the form:
where a=1 and b=11
The function is
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Answer:
The second choice: graph begins in the second quadrant near the axis and increases slowly while crossing the ordered pair 3, 1. The graph then begins to increase quickly throughout the first quadrant.
Explanation:
1) As per the set of choices, the function is:
2) Therefore it is an exponential function with these characteristics:
- Since, the bases is greater than 1 (2), the function is growing in all the domain.
- The domain is all real numbers (- ∞, ∞).
- To know where the function starts, calculate the limit of f(x) as x trends to negative infinity:
That means that the range is y > 0, and so the graph starts in the second quadrant.
- You can find the y-intersection making x = 0, which is 2⁰ ⁻ ³ = 2 ⁻³ = 1/8 = 0.25. So, the graph cross the y axis at y = 0.25.
- That tells you, that the function increases slowly, at least, until that point (0, 0.25).
- The other bullet point is when x = 3: 2 ³ ⁻ ³ = 2⁰ = 1. Therefore, the graph passes through the point (3, 1).
- From that point, the function starts to increase rapidly (since it is exponential).
- Those are the characteristics given by the second choice: graph begins in the second quadrant near the x-axis and increases slowly while crossing the ordered pair 3, 1. The graph then begins to increase quickly throughout the first quadrant.
You surely will find useful to watch the graph that I have attached.
