Question

5. Consider the equation x2 + 3x + 2y = 6 a) (10 points) Use implicit differentiation to find dy dx b) 6 points) Find an equation for the tangent line to the curve described by the equation in part a) at the point (-1,2).

please answer both parts of question 5

Answer 1

z? + 3.c + 2y2 = 6 Remember: daz") dx n.7-1 d(C) do 0 So, 2.1 +3* 1+2*2*y* dy dc 0 dy dac (2x + 3) 4y

B.

Now tangent line of curve at point P (x1, y1) will be given by:

y - y1 = m*(x - x1)

where m = slope of tangent line = dy/dx at point P

Given P (-1, 2), So

m = -(2*(-1) + 3)/(4*2) = -1/8

So,

y - 2 = (-1/8)*(x - (-1))

8y - 16 = -x - 1

x + 8y - 15 = 0

Let me know if you've any query.