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

The tangent of f(x)=2x^2+x-3,\at x=-1 is y=-3x-5

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

tangent of

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

y = - 3 x - 5

Solution steps

Find the tangent point:

(- 1, - 2)

Find the slope of

f (x) = 2 x^{2} + x - 3 : \frac{df ( x )}{d x} = 4 x + 1

Compute the slope of

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

x = - 1 : m = - 3

Find the line with slope m=

- 3

and passing through

(- 1, - 2) : y = - 3 x - 5

y = - 3 x - 5

y^{^{'}} = (t + y)^{2}

2 x, 3 x^{2}, 0, 1

\frac{d y}{d t} = \frac{t e ^{t}}{y 6 + y ^{2}}

e^{\frac{x}{2}}

x \to \infty lim (1 - 2^{x})

What is the tangent of f(x)=2x^2+x-3,\at x=-1 ?
The tangent of f(x)=2x^2+x-3,\at x=-1 is y=-3x-5