Please Help ASAP!! Marking Brainliest!! Divide
id="TexFormula1" title="12x^2 + 4x^3 + 18 + 16x" alt="12x^2 + 4x^3 + 18 + 16x" align="absmiddle" class="latex-formula"> by 
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Answer:
- lower right (see attachment 1)
- 39
- linear/exponential (see attachment 2)
Step-by-step explanation:
<h3>1.</h3>An exponential function is one that increases or decreases at a rate that is proportional to its value. That is, if it is increasing, it does so at an increasing rate. If it is decreasing, it does so at a decreasing rate.
The "base" of the function is the number in parentheses when the function is written as it is here. The multiplier is the number to the left of these parentheses, and the exponent is the small number or expression to the upper right of the parentheses.
If the "base" is less than 1, then the function will be decreasing. If there is no added constant, it will decrease toward zero. If there is an added constant (as here, -1), then the function will decrease toward that value. The only graph showing a function decreasing toward -1 is the one at lower right (attachment 1).
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<em>Note about the multiplier</em>
If the multiplier is negative (not the case here), then the graph will be reflected across the x-axis and the descriptors "increasing" and "decreasing" get reversed. A graph that would go down to the right with a positive multiplier goes up to the right with a negative multiplier.
What you can <em>always</em> say is that when the base is less than 1, the graph is steep on the left and flat on the right. For a base greater than 1, the reverse is true.
<h3>2.</h3>The average rate of change of a function between x1 and x2 is given by the formula ...
... average rate of change = (f(x2) -f(x1))/(x2 -x1)
Here, you have x2 = 10, so f(x2) = 3^(10/2) = 3^5 = 243; and x1 = 4, so f(x1) = 3^(4/2) = 3^2 = 9.
Then the average rate of change is ...
... average rate of change = (243 -9)/(10 -4) = 234/6 = 39 . . . flowers/day
<h3>3.</h3>When faced with determining the sort of function you have, first of all check the given x-values. If they are evenly spaced, as here, then you are in luck.
Next look at the differences of y-values. These are the "first differences". For the first table, they are ...
- 7 - 5.5 = 1.5
- 8.5 - 7 = 1.5 . . . . same as the other one
When the first differences are constant, as here, the function is linear.
In the second table, the first differences are decidedly <u>not</u> constant, so the function is not linear. Another test you can do is to see if the ratios of y-values are a constant. Here, those ratios are ...
- 8/1 = 8
- 64/8 = 8
These ratios are constant, so the function is exponential.
