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

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

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

tangent of

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

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

Solution steps

Find the slope of

y = 4 x + 33 : \frac{d y}{d x} = \frac{2}{4 x + 33}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}

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

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

\int \frac{1}{cot ( 3 x )} d x

y^{^{'}} = \frac{x ^{2} + 3}{y}, y (0) = 1

x e^{x}

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