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

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

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

tangent of

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

y = - 3 x - 3

Solution steps

Find the tangent point:

(- 1, 0)

Find the slope of

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

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 = - 3

Find the line with slope m=

- 3

and passing through

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

y = - 3 x - 3

\frac{d y}{d x} + y = y^{2}

\int x \cdot e^{1 - x^{2}} d x

\int \frac{( e ^{x} + 1 ) ^{2}}{e ^{x}} d x

x^{l n (x)}

n = 1 \sum \infty \frac{3}{n ^{2} + 1}

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