What is the tangent of y=sqrt(x+1),(8,3) ?

The tangent of y=sqrt(x+1),(8,3) is y= 1/6 x+5/3

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tangent of y=sqrt(x+1),(8,3)

tangent of

y = x + 1, (8, 3)

y = \frac{1}{6} x + \frac{5}{3}

Solution steps

Find the slope of

y = x + 1 : \frac{d y}{d x} = \frac{1}{2 x + 1}

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}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(8, 3) : y = \frac{1}{6} x + \frac{5}{3}

y = \frac{1}{6} x + \frac{5}{3}

f (x) = x^{4} - 2 x^{3} + x - 1

\frac{d}{d t} (e^{r (t) t})

- x^{2} + 9 x, x^{2} - 5 x

y = 3 - 5 x

\int_{4}^{9} \frac{1}{8 - x} d x

What is the tangent of y=sqrt(x+1),(8,3) ?
The tangent of y=sqrt(x+1),(8,3) is y= 1/6 x+5/3