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Maslowich
3 years ago
6

I want solution 1 and 2 with the steps to know the final answer

Mathematics
1 answer:
Delicious77 [7]3 years ago
4 0
Question 1:

Slope = 1/5

y = mx + c 
y = 1/5 x + c
at point (5, -1)
-1 = 1/5 (5) + c
- 1= 1 + c
c = - 2

y = 1/5x - 2
5y = x - 10

Question 2:
slope = (9-5)/(3-1)
Slope = 2

y = mx + c
y = 2x + c

at point (1, 5)
5 = 2(1) + c
c = 5 - 2
c = 3

y = 2x + 3


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Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.

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so either
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But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well. 
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