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

The tangent of f(x)=-3x^2-2x^2-1,\at x=-1 is y=10x+4

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

tangent of

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

y = 10 x + 4

Solution steps

Find the tangent point:

(- 1, - 6)

Find the slope of

f (x) = - 3 x^{2} - 2 x^{2} - 1 : \frac{df ( x )}{d x} = - 10 x

Compute the slope of

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

x = - 1 : m = 10

Find the line with slope m=

10

and passing through

(- 1, - 6) : y = 10 x + 4

y = 10 x + 4

\int \frac{cos ( x )}{5 + sin ( x )} d x

11 t^{2} + 9

\int_{0}^{3} \frac{1}{x ^{2} - 6 x + 5} d x

\frac{6 x}{( x ^{2} + 3 ) ^{2}}

y = 7^{x + a r c c o s (x)}

What is the tangent of f(x)=-3x^2-2x^2-1,\at x=-1 ?
The tangent of f(x)=-3x^2-2x^2-1,\at x=-1 is y=10x+4