What is the tangent of f(x)=2x^2+1,(-1,3) ?

The tangent of f(x)=2x^2+1,(-1,3) is y=-4x-1

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tangent of f(x)=2x^2+1,(-1,3)

tangent of

f (x) = 2 x^{2} + 1, (- 1, 3)

y = - 4 x - 1

Solution steps

Find the slope of

f (x) = 2 x^{2} + 1 : \frac{df ( x )}{d x} = 4 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 4

Find the line with slope m=

- 4

and passing through

(- 1, 3) : y = - 4 x - 1

y = - 4 x - 1

f (x) = \frac{e ^{x}}{9 x ^{2} + 2}

\int \frac{1}{u ^{2}} d u

\int 4 e^{- 3 x} d x

\frac{d}{d x} (x^{\frac{1}{8}})

f (x) = \frac{x}{6}

What is the tangent of f(x)=2x^2+1,(-1,3) ?
The tangent of f(x)=2x^2+1,(-1,3) is y=-4x-1