What is the tangent of y=(x^2+2)/(x+8),(3,1) ?

The tangent of y=(x^2+2)/(x+8),(3,1) is y= 5/11 x-4/11

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tangent of y=(x^2+2)/(x+8),(3,1)

tangent of

y = \frac{x ^{2} + 2}{x + 8}, (3, 1)

y = \frac{5}{11} x - \frac{4}{11}

Solution steps

Find the slope of

y = \frac{x ^{2} + 2}{x + 8} : \frac{d y}{d x} = \frac{x ^{2} + 16 x - 2}{( x + 8 ) ^{2}}

Compute the slope of

y = \frac{x ^{2} + 2}{x + 8}

(3, 1) : m = \frac{5}{11}

Find the line with slope m=

\frac{5}{11}

and passing through

(3, 1) : y = \frac{5}{11} x - \frac{4}{11}

y = \frac{5}{11} x - \frac{4}{11}

\int 2 e^{3 x - 5} d x

x \to π lim (tan (\frac{5 x}{3}))

\int - 2 + 36 x - 12 x^{2} d x

x \to 81 lim (x^{\frac{3}{2}})

\int \frac{- 9 + 9 sin ( x )}{cos ( x )} d x

What is the tangent of y=(x^2+2)/(x+8),(3,1) ?
The tangent of y=(x^2+2)/(x+8),(3,1) is y= 5/11 x-4/11