What is the tangent of f(x)=6x^2,\at x=10 ?

The tangent of f(x)=6x^2,\at x=10 is y=120x-600

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tangent of f(x)=6x^2,\at x=10

tangent of

f (x) = 6 x^{2}, at x = 10

y = 120 x - 600

Solution steps

Find the tangent point:

(10, 600)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 120

Find the line with slope m=

120

and passing through

(10, 600) : y = 120 x - 600

y = 120 x - 600

\frac{48}{( s + 2 ) ( s ^{2} + 16 )}

x \to 1 lim (3 x^{4} - 3 x^{2} + 6)

\int_{0}^{2} \frac{x}{4 - x ^{2}} d x

x = y^{2} - 12, x = y

e^{3 x^{2}}

What is the tangent of f(x)=6x^2,\at x=10 ?
The tangent of f(x)=6x^2,\at x=10 is y=120x-600