What is the tangent of f(x)=8x^2,\at x=10 ?

The tangent of f(x)=8x^2,\at x=10 is y=160x-800

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tangent of f(x)=8x^2,\at x=10

tangent of

f (x) = 8 x^{2}, at x = 10

y = 160 x - 800

Solution steps

Find the tangent point:

(10, 800)

Find the slope of

f (x) = 8 x^{2} : \frac{df ( x )}{d x} = 16 x

Compute the slope of

f (x) = 8 x^{2}

x = 10 : m = 160

Find the line with slope m=

160

and passing through

(10, 800) : y = 160 x - 800

y = 160 x - 800

\int \frac{5 - x ^{2}}{x ^{2}} d x

\int 1 + tan^{2} (x) - 2 cos (x) d x

\frac{\partial}{\partial r} (\frac{π}{3} h^{2} (3 r - h))

e^{x} y d y = (e^{- y} + e^{2 x - y}) d x

n = 1 \sum \infty n e^{5 n}

What is the tangent of f(x)=8x^2,\at x=10 ?
The tangent of f(x)=8x^2,\at x=10 is y=160x-800