What is the tangent of f(x)=(x^2+8)/(x+28),(5,1) ?

The tangent of f(x)=(x^2+8)/(x+28),(5,1) is y= 3/11 x-4/11

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tangent of f(x)=(x^2+8)/(x+28),(5,1)

tangent of

f (x) = \frac{x ^{2} + 8}{x + 28}, (5, 1)

y = \frac{3}{11} x - \frac{4}{11}

Solution steps

Find the slope of

f (x) = \frac{x ^{2} + 8}{x + 28} : \frac{df ( x )}{d x} = \frac{x ^{2} + 56 x - 8}{( x + 28 ) ^{2}}

Compute the slope of

f (x) = \frac{x ^{2} + 8}{x + 28}

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

Find the line with slope m=

\frac{3}{11}

and passing through

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

y = \frac{3}{11} x - \frac{4}{11}

(4, - 9), (2, - 2)

\int x^{m - 1} d x

y^{^{'}} = \frac{1}{1 - x ^{2}}

\int y e^{3 y^{2}} d y

\frac{d}{d x} (I n x^{2})

What is the tangent of f(x)=(x^2+8)/(x+28),(5,1) ?
The tangent of f(x)=(x^2+8)/(x+28),(5,1) is y= 3/11 x-4/11