Find the slope of the line tangent to the graph of x²+2xy² +3y=31 at the point (2, -3)
1 answer:
9514 1404 393
Answer:
22/21
Step-by-step explanation:
Taking the derivative, we have ...
2x·dx +2y²·dx +4xy·dy +3·dy = 0
dx(2x +2y²) +dy(4xy+3) = 0
At the given point, this is ...
dx(2·2 +2·(-3)²) +dy(4·2·(-3) +3) = 0
22dx -21dy = 0
dy/dx = 22/21
The slope of the tangent at the point of interest is 22/21.
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Answer:
x=19, y=-5
Step-by-step explanation:
Add the equations in order to solve for the first variable. Plug this value into the other equations in order to solve for the remaining variables.
Point Form:
(19, -5)
Equation Form:
x=19, y=-5
Hope this helps.
