What is the tangent of f(x)=-1-9x^2,\at x=3 ?

The tangent of f(x)=-1-9x^2,\at x=3 is y=-54x+80

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tangent of f(x)=-1-9x^2,\at x=3

tangent of

f (x) = - 1 - 9 x^{2}, at x = 3

y = - 54 x + 80

Solution steps

Find the tangent point:

(3, - 82)

Find the slope of

f (x) = - 1 - 9 x^{2} : \frac{df ( x )}{d x} = - 18 x

Compute the slope of

f (x) = - 1 - 9 x^{2}

x = 3 : m = - 54

Find the line with slope m=

- 54

and passing through

(3, - 82) : y = - 54 x + 80

y = - 54 x + 80

r^{2} - x^{2}

x^{2}, 4

\int \frac{1}{x x + 6} d x

\int \frac{( x ^{a} - x ^{b} ) ^{2}}{x} d x

x \to 0 lim (\frac{e ^{4 x} - 1}{x})

What is the tangent of f(x)=-1-9x^2,\at x=3 ?
The tangent of f(x)=-1-9x^2,\at x=3 is y=-54x+80