Use the diagrams to answer the problems,
1 answer:
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Answer:
B
Step-by-step explanation:
Let's create an equation in point-slope form, which is: , where
is the point and m is the slope.
Here, our point is (-2, 7) and our slope is m = -5 so plug these in:
Now, if (a, 2) lies on this line, then it should satisfy the equation. So plug in a for x and 2 for y:
2 - 7 = -5(a + 2)
-5 = -5(a + 2)
Divide by -5:
1 = a + 2
Subtract 2 from both sides:
a = -1
Thus, the answer is B.
Hope this helps!
Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0
