What is the tangent of f(x)=sqrt(4x+33),(4,7) ?

The tangent of f(x)=sqrt(4x+33),(4,7) is y= 2/7 x+41/7

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tangent of f(x)=sqrt(4x+33),(4,7)

tangent of

f (x) = 4 x + 33, (4, 7)

y = \frac{2}{7} x + \frac{41}{7}

Solution steps

Find the slope of

f (x) = 4 x + 33 : \frac{df ( x )}{d x} = \frac{2}{4 x + 33}

Compute the slope of

f (x) = 4 x + 33

(4, 7) : m = \frac{2}{7}

Find the line with slope m=

\frac{2}{7}

and passing through

(4, 7) : y = \frac{2}{7} x + \frac{41}{7}

y = \frac{2}{7} x + \frac{41}{7}

\frac{\partial}{\partial y} (sin (x y^{2}))

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

\int \frac{y}{y ^{2} - 1} d y

sin (cos (tan (x)))

\frac{\partial}{\partial y} (\frac{x}{y + 2 z})

What is the tangent of f(x)=sqrt(4x+33),(4,7) ?
The tangent of f(x)=sqrt(4x+33),(4,7) is y= 2/7 x+41/7