What is the tangent of y=sqrt(3x+7),(6,5) ?

The tangent of y=sqrt(3x+7),(6,5) is y= 3/10 x+16/5

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tangent of y=sqrt(3x+7),(6,5)

tangent of

y = 3 x + 7, (6, 5)

y = \frac{3}{10} x + \frac{16}{5}

Solution steps

Find the slope of

y = 3 x + 7 : \frac{d y}{d x} = \frac{3}{2 3 x + 7}

Compute the slope of

y = 3 x + 7

(6, 5) : m = \frac{3}{10}

Find the line with slope m=

\frac{3}{10}

and passing through

(6, 5) : y = \frac{3}{10} x + \frac{16}{5}

y = \frac{3}{10} x + \frac{16}{5}

y = 2 x^{3} + 3 x^{2} - 12 x + 8

\int 13 x^{3} e^{x} d x

\frac{( x + 1 )}{x + 5}

\int \frac{5 x ^{2} + 3 x + 5}{( x ^{2} + 1 ) ^{2}} d x

\frac{\partial}{\partial x} (12 x + 2 y - 3 x^{2} - y^{2})

What is the tangent of y=sqrt(3x+7),(6,5) ?
The tangent of y=sqrt(3x+7),(6,5) is y= 3/10 x+16/5