What is the tangent of f(x)=-6x^2-10x,\at x=1 ?

The tangent of f(x)=-6x^2-10x,\at x=1 is y=-22x+6

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tangent of f(x)=-6x^2-10x,\at x=1

tangent of

f (x) = - 6 x^{2} - 10 x, at x = 1

y = - 22 x + 6

Solution steps

Find the tangent point:

(1, - 16)

Find the slope of

f (x) = - 6 x^{2} - 10 x : \frac{df ( x )}{d x} = - 12 x - 10

Compute the slope of

f (x) = - 6 x^{2} - 10 x

x = 1 : m = - 22

Find the line with slope m=

- 22

and passing through

(1, - 16) : y = - 22 x + 6

y = - 22 x + 6

\frac{t ^{2} + 1}{t ^{2} - 1}

y = x (2 x + 3)^{5}

\frac{\partial}{\partial x} (arcsin (x - y))

\int (x^{n}) d x

x e^{- x^{2}}

What is the tangent of f(x)=-6x^2-10x,\at x=1 ?
The tangent of f(x)=-6x^2-10x,\at x=1 is y=-22x+6