What is the tangent of y=5x^2,(4,80) ?

The tangent of y=5x^2,(4,80) is y=40x-80

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tangent of y=5x^2,(4,80)

tangent of

y = 5 x^{2}, (4, 80)

y = 40 x - 80

Solution steps

Find the slope of

y = 5 x^{2} : \frac{d y}{d x} = 10 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 = 40

Find the line with slope m=

40

and passing through

(4, 80) : y = 40 x - 80

y = 40 x - 80

x \to - \infty lim (- ln (x))

\frac{d}{d x} (\frac{x}{x ^{2} + 1})

f^{^{'}} (x) = - 2 ln (x) + x^{2} - 4 + e^{x} + x^{e}

f (x) = x^{2} - 6 x + 8, g (x) = - 2 x + 8

y^{^{''}} + 8 y^{^{'}} + 16 = 0