What is the tangent of y=sqrt(x)(x-8),(16,32) ?

The tangent of y=sqrt(x)(x-8),(16,32) is y=5x-48

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tangent of y=sqrt(x)(x-8),(16,32)

tangent of

y = x (x - 8), (16, 32)

y = 5 x - 48

Solution steps

Find the slope of

y = x (x - 8) : \frac{d y}{d x} = \frac{3 x - 8}{2 x}

Compute the slope of

y = x (x - 8)

(16, 32) : m = 5

Find the line with slope m=

5

and passing through

(16, 32) : y = 5 x - 48

y = 5 x - 48

\int_{0}^{1} x^{2} - 2 x + 1 d x

f (x) = 3 x + 1, (5, 4)

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

t^{2} y^{^{''}} + 2 t y^{^{'}} - 6 y = 0

\int \frac{1}{3 + t} d t

What is the tangent of y=sqrt(x)(x-8),(16,32) ?
The tangent of y=sqrt(x)(x-8),(16,32) is y=5x-48