What is the tangent of ((x^2+32sqrt(x)))/8 ,\at x=4 ?

The tangent of ((x^2+32sqrt(x)))/8 ,\at x=4 is y=2x+2

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tangent of ((x^2+32sqrt(x)))/8 ,\at x=4

tangent of

\frac{( x ^{2} + 32 x )}{8}, at x = 4

y = 2 x + 2

Solution steps

Find the tangent point:

(4, 10)

Find the slope of

y = \frac{x ^{2} + 32 x}{8} : \frac{d y}{d x} = \frac{1}{8} (2 x + \frac{16}{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 = 2

Find the line with slope m=

2

and passing through

(4, 10) : y = 2 x + 2

y = 2 x + 2

What is the tangent of ((x^2+32sqrt(x)))/8 ,\at x=4 ?
The tangent of ((x^2+32sqrt(x)))/8 ,\at x=4 is y=2x+2