How do you find the equation of tangent line to the curve $f (x) = 2 x^{2}$ $f(x)=2x^2$ at x=-1?

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Answer 1 · 2016-03-28T14:05:18

Equation of tangent is $4 x + y + 2 = 0$ $4x+y+2=0$

At $x = 1$ $x=1$ , $f (x) = 2 \times {(- 1)}^{2} = 2$ $f(x)=2xx(-1)^2=2$ hence tangent passes through $(- 1, 2)$ $(-1,2)$

Slope of the line is given by value of derivative and as

$f'(x)=4x$ , slope of tangent at $(-1,2)$ will be $4xx(-1)=-4$

Hence equation of tangent is

$(y-2)=-4(x+1)$ or

$4x+y+2=0$

How do you find the equation of tangent line to the curve f(x)=2x^2f(x)=2x2f(x)=2x^2 at x=-1?