What is the tangent of sqrt(3x+25),\at x=-3 ?

The tangent of sqrt(3x+25),\at x=-3 is y= 3/8 x+41/8

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tangent of sqrt(3x+25),\at x=-3

tangent of

3 x + 25, at x = - 3

y = \frac{3}{8} x + \frac{41}{8}

Solution steps

Find the tangent point:

(- 3, 4)

Find the slope of

y = 3 x + 25 : \frac{d y}{d x} = \frac{3}{2 3 x + 25}

Compute the slope of

y = 3 x + 25

x = - 3 : m = \frac{3}{8}

Find the line with slope m=

\frac{3}{8}

and passing through

(- 3, 4) : y = \frac{3}{8} x + \frac{41}{8}

y = \frac{3}{8} x + \frac{41}{8}

\int (2 x - 1) ln (5 x) d x

\frac{d P}{d t} = P t

\int x^{4} ln (2 x) d x

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

f (b) = \frac{b}{z + n}

What is the tangent of sqrt(3x+25),\at x=-3 ?
The tangent of sqrt(3x+25),\at x=-3 is y= 3/8 x+41/8