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

The tangent of f(x)=-3-5x^2,\at x=3 is y=-30x+42

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tangent of f(x)=-3-5x^2,\at x=3

tangent of

f (x) = - 3 - 5 x^{2}, at x = 3

y = - 30 x + 42

Solution steps

Find the tangent point:

(3, - 48)

Find the slope of

f (x) = - 3 - 5 x^{2} : \frac{df ( x )}{d x} = - 10 x

Compute the slope of

f (x) = - 3 - 5 x^{2}

x = 3 : m = - 30

Find the line with slope m=

- 30

and passing through

(3, - 48) : y = - 30 x + 42

y = - 30 x + 42

\int \frac{1}{2 + 3 x - 2 x ^{2}} d x

\frac{d y}{d x} = \frac{( x - y )}{x}

e^{\frac{a ^{2}}{2 a + 1}}

\frac{d}{d x} (2 y - \frac{1}{x} + cos (3 x))

\int x^{2} \cdot cos (mπ x) d x

What is the tangent of f(x)=-3-5x^2,\at x=3 ?
The tangent of f(x)=-3-5x^2,\at x=3 is y=-30x+42