What is the tangent of y=sqrt(5x+20),\at x=1 ?

The tangent of y=sqrt(5x+20),\at x=1 is y= 1/2 x+9/2

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tangent of y=sqrt(5x+20),\at x=1

tangent of

y = 5 x + 20, at x = 1

y = \frac{1}{2} x + \frac{9}{2}

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

y = 5 x + 20 : \frac{d y}{d x} = \frac{5}{2 5 x + 20}

Compute the slope of

y = 5 x + 20

x = 1 : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(1, 5) : y = \frac{1}{2} x + \frac{9}{2}

y = \frac{1}{2} x + \frac{9}{2}

f (x) = 16 x + 3

2 x^{2} y y^{^{'}} + 1 + 2 x y^{2} = 0

\int_{- 2}^{1} x ((4 - x^{2}) - (x + 2)) d x

d x = (e^{t^{2}} + 2 x t) d t

\int 1 3^{x} d x

What is the tangent of y=sqrt(5x+20),\at x=1 ?
The tangent of y=sqrt(5x+20),\at x=1 is y= 1/2 x+9/2