What is the tangent of y=sqrt(4x+48),(4,8) ?

The tangent of y=sqrt(4x+48),(4,8) is y= 1/4 x+7

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tangent of y=sqrt(4x+48),(4,8)

tangent of

y = 4 x + 48, (4, 8)

y = \frac{1}{4} x + 7

Solution steps

Find the slope of

y = 4 x + 48 : \frac{d y}{d x} = \frac{1}{x + 12}

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{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(4, 8) : y = \frac{1}{4} x + 7

y = \frac{1}{4} x + 7

\frac{74}{s ^{2} + 1}

\int 3 cos^{3} (3 x) d x

\int_{4}^{5} x x - 4 d x

\frac{d}{d x} (- 0.5 x^{2} + 56 x + 20)

\frac{( 4 - y )}{( 2 ^{y} + 3 )}

What is the tangent of y=sqrt(4x+48),(4,8) ?
The tangent of y=sqrt(4x+48),(4,8) is y= 1/4 x+7