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

The tangent of y=sqrt(x-8) is y= 1/(2sqrt(a_{0)-8)}x+sqrt(a_{0)-8}-(a_{0})/(2sqrt(a_{0)-8)}

Popular >

tangent of y=sqrt(x-8)

tangent of

y = x - 8

y = \frac{1}{2 a _{0} - 8} x + a_{0} - 8 - \frac{a _{0}}{2 a _{0} - 8}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0} - 8)

Find the slope of

y = x - 8 : \frac{d y}{d x} = \frac{1}{2 x - 8}

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{1}{2 a _{0} - 8}

Find the line with slope m=

\frac{1}{2 a _{0} - 8}

and passing through

(a_{0}, a_{0} - 8) : y = \frac{1}{2 a _{0} - 8} x + a_{0} - 8 - \frac{a _{0}}{2 a _{0} - 8}

y = \frac{1}{2 a _{0} - 8} x + a_{0} - 8 - \frac{a _{0}}{2 a _{0} - 8}

n = 4 \sum \infty \frac{n}{n - 3}

\frac{d y}{d x} - e^{2 x + y} = 0

\frac{d}{d x} (a x^{3} + b x^{2} + c x)

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

\frac{\partial ^{2}}{\partial x \partial y} (x^{2} - e^{y^{2}})

What is the tangent of y=sqrt(x-8) ?
The tangent of y=sqrt(x-8) is y= 1/(2sqrt(a_{0)-8)}x+sqrt(a_{0)-8}-(a_{0})/(2sqrt(a_{0)-8)}