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

The tangent of f(x)=5x^2-5x,\at x=3 is y=25x-45

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

tangent of

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

y = 25 x - 45

Solution steps

Find the tangent point:

(3, 30)

Find the slope of

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

Compute the slope of

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

x = 3 : m = 25

Find the line with slope m=

25

and passing through

(3, 30) : y = 25 x - 45

y = 25 x - 45

\int \frac{arctan ( \frac{x}{2} )}{4 + x ^{2}} d x

\frac{d}{d x} (x^{3} - x + 1)

\int (cosh (x))^{3} d x

\frac{\partial}{\partial x} (x^{\frac{1}{2}} + \frac{Y ( x )}{2})

f (x) = \frac{x ^{2} - 1}{x ^{2} + x + 1}

What is the tangent of f(x)=5x^2-5x,\at x=3 ?
The tangent of f(x)=5x^2-5x,\at x=3 is y=25x-45