How to solve q =r/2(s+t) for t
1 answer:
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Answer:
(a) y = 9/4x -1/4
(b) (-2, -2), (-2, 2)
(c) see below
Step-by-step explanation:
(a) The derivative can be found from ...
2y·y' = 3x² +6x
At (x, y) = (1, 2), this is ...
4y' = 9
y' = 9/4
so the equation of the tangent is ...
y = (9/4)(x -1) +2
y = (9/4)x -1/4
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(b) At y' = 0, we have ...
0 = 3x² +6x = 3x(x +2)
This says y' = 0 at x=0 or x=-2. (x = 0 is an extraneous solution.)
At x = -2, we have ...
y² = (-2)³ +3(-2)² = -8+12 = 4
y = ±2
So, the horizontal tangent points are (x, y) = (-2, ±2).
