What is the tangent of f(x)=1-6x^2,(2,-23) ?

The tangent of f(x)=1-6x^2,(2,-23) is y=-24x+25

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tangent of f(x)=1-6x^2,(2,-23)

tangent of

f (x) = 1 - 6 x^{2}, (2, - 23)

y = - 24 x + 25

Solution steps

Find the slope of

f (x) = 1 - 6 x^{2} : \frac{df ( x )}{d x} = - 12 x

Compute the slope of

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

(2, - 23) : m = - 24

Find the line with slope m=

- 24

and passing through

(2, - 23) : y = - 24 x + 25

y = - 24 x + 25

\frac{\partial}{\partial x} (x y + y^{2})

x^{3} + y^{3} - 9 x y = 0, (4, 2)

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

\int \frac{( 4 x ^{m} - 3 x ^{n} ) ^{3}}{x} d x

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

What is the tangent of f(x)=1-6x^2,(2,-23) ?
The tangent of f(x)=1-6x^2,(2,-23) is y=-24x+25