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toofehblf

Answered question

2022-02-11

How do you find the slope of the line tangent to the curve

3 x^{2} - 2 x y + y = 11

at the point (1,-2)?

monogamab0f

Beginner2022-02-12Added 9 answers

We have that y=y(x) hence the tangent is
y-y(1)=y'(1)(x-1)
we know that y(1)=-2 and we need to find y'(1).
The derivative of the given equation is

d \frac{3 x^{2} - 2 x y + y - 11}{d x} = 6 x - 2 y - 2 x \frac{d y}{d x} + \frac{d y}{d y} = 0

For x=1, y=-2 we get

6 \cdot 1 - 2 \cdot (- 2) - 2 \cdot 1 \frac{d y}{d x} + \frac{d y}{d x} = 0 \Rightarrow 10 - \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = 10

Hence the slope is y'(1)=10 and the tangent is

y - 1 = 10 \cdot (x - 1) \Rightarrow y = 10 x - 9

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