What is the tangent of f(x)=sqrt(3x+10),(2,4) ?

The tangent of f(x)=sqrt(3x+10),(2,4) is y= 3/8 x+13/4

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tangent of f(x)=sqrt(3x+10),(2,4)

tangent of

f (x) = 3 x + 10, (2, 4)

y = \frac{3}{8} x + \frac{13}{4}

Solution steps

Find the slope of

f (x) = 3 x + 10 : \frac{df ( x )}{d x} = \frac{3}{2 3 x + 10}

Compute the slope of

f (x) = 3 x + 10

(2, 4) : m = \frac{3}{8}

Find the line with slope m=

\frac{3}{8}

and passing through

(2, 4) : y = \frac{3}{8} x + \frac{13}{4}

y = \frac{3}{8} x + \frac{13}{4}

\int \frac{26}{( x - 1 ) ( x ^{2} + 25 )} d x

(x^{2} - 7 x + 4) 25 - x^{2}

\int \frac{x + 1}{3} d x

x y^{2} d x + (1 + x^{2} y) d y = 0

\int e^{(x^{2} - l n (x))} d x

What is the tangent of f(x)=sqrt(3x+10),(2,4) ?
The tangent of f(x)=sqrt(3x+10),(2,4) is y= 3/8 x+13/4