What is the tangent of sqrt(3x-5) ?

The tangent of sqrt(3x-5) is y= 3/(2sqrt(3a_{0)-5)}x+sqrt(3a_{0)-5}-(3a_{0})/(2sqrt(3a_{0)-5)}

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tangent of sqrt(3x-5)

tangent of

3 x - 5

y = \frac{3}{2 3 a _{0} - 5} x + 3 a_{0} - 5 - \frac{3 a _{0}}{2 3 a _{0} - 5}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0} - 5)

Find the slope of

y = 3 x - 5 : \frac{d y}{d x} = \frac{3}{2 3 x - 5}

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 = \frac{3}{2 3 a _{0} - 5}

Find the line with slope m=

\frac{3}{2 3 a _{0} - 5}

and passing through

(a_{0}, 3 a_{0} - 5) : y = \frac{3}{2 3 a _{0} - 5} x + 3 a_{0} - 5 - \frac{3 a _{0}}{2 3 a _{0} - 5}

y = \frac{3}{2 3 a _{0} - 5} x + 3 a_{0} - 5 - \frac{3 a _{0}}{2 3 a _{0} - 5}

What is the tangent of sqrt(3x-5) ?
The tangent of sqrt(3x-5) is y= 3/(2sqrt(3a_{0)-5)}x+sqrt(3a_{0)-5}-(3a_{0})/(2sqrt(3a_{0)-5)}