No local extrema i think?
Answer:
(5, 7)
Step-by-step explanation:
before the translation, A is (2, 5)
by adding 3 to A and 2 to y, we get (5, 7)
Hello.
Simply plug -1 into 2x+2.
f(x) = 2x +2
f(-1) = 2 (-1) +2
f (-1)= -2+2
f(-1) = 0
Hope this helped.
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Disclaimer:
Always double check your answer with a reliable source, as mistakes can be made.
Answer:
16.5 square units
Step-by-step explanation:
You are expected to integrate the function between x=1 and x=4:
![\displaystyle\text{area}=\int_1^4{(5x-x^2)}\,dx=\left.\left(\dfrac{5}{2}x^2-\dfrac{1}{3}x^3\right)\right|_{x=1}^{x=4}\\\\=\dfrac{5(4^2-1^2)}{2}-\dfrac{4^3-1^3}{3}=37.5-21=\boxed{16.5}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ctext%7Barea%7D%3D%5Cint_1%5E4%7B%285x-x%5E2%29%7D%5C%2Cdx%3D%5Cleft.%5Cleft%28%5Cdfrac%7B5%7D%7B2%7Dx%5E2-%5Cdfrac%7B1%7D%7B3%7Dx%5E3%5Cright%29%5Cright%7C_%7Bx%3D1%7D%5E%7Bx%3D4%7D%5C%5C%5C%5C%3D%5Cdfrac%7B5%284%5E2-1%5E2%29%7D%7B2%7D-%5Cdfrac%7B4%5E3-1%5E3%7D%7B3%7D%3D37.5-21%3D%5Cboxed%7B16.5%7D)
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<em>Additional comment</em>
If you're aware that the area inside a (symmetrical) parabola is 2/3 of the area of the enclosing rectangle, you can compute the desired area as follows.
The parabolic curve is 4-1 = 3 units wide between x=1 and x=4. It extends upward 2.25 units from y=4 to y=6.25, so the enclosing rectangle is 3×2.25 = 6.75 square units. 2/3 of that area is (2/3)(6.75) = 4.5 square units.
This region sits on top of a rectangle 3 units wide and 4 units high, so the total area under the parabolic curve is ...
area = 4.5 +3×4 = 16.5 . . . square units
Answer:
10/11
Step-by-step explanation: