What is the tangent of y=sqrt(3x+72),(3,9) ?

The tangent of y=sqrt(3x+72),(3,9) is y= 1/6 x+17/2

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tangent of y=sqrt(3x+72),(3,9)

tangent of

y = 3 x + 72, (3, 9)

y = \frac{1}{6} x + \frac{17}{2}

Solution steps

Find the slope of

y = 3 x + 72 : \frac{d y}{d x} = \frac{3}{2 3 x + 72}

Compute the slope of

y = 3 x + 72

(3, 9) : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(3, 9) : y = \frac{1}{6} x + \frac{17}{2}

y = \frac{1}{6} x + \frac{17}{2}

y = \frac{7 x}{8}

x \to 0 lim (\frac{x ^{2}}{e ^{x} - 1})

\int 3 sec (x) d x

\frac{\partial}{\partial x} (a^{2})

\int (x - 2)^{\frac{3}{2}} d x

What is the tangent of y=sqrt(3x+72),(3,9) ?
The tangent of y=sqrt(3x+72),(3,9) is y= 1/6 x+17/2