What is the tangent of f(x)=sqrt(x^2+16),\at x=-3 ?

The tangent of f(x)=sqrt(x^2+16),\at x=-3 is y=-3/5 x+16/5

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tangent of f(x)=sqrt(x^2+16),\at x=-3

tangent of

f (x) = x^{2} + 16, at x = - 3

y = - \frac{3}{5} x + \frac{16}{5}

Solution steps

Find the tangent point:

(- 3, 5)

Find the slope of

f (x) = x^{2} + 16 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 16}

Compute the slope of

f (x) = x^{2} + 16

x = - 3 : m = - \frac{3}{5}

Find the line with slope m=

- \frac{3}{5}

and passing through

(- 3, 5) : y = - \frac{3}{5} x + \frac{16}{5}

y = - \frac{3}{5} x + \frac{16}{5}

(x + 5) cos (x)

y^{^{''}} + 4 y^{^{'}} + 9 y = 0

\int \frac{x ^{2}}{196 - x ^{2}} d x

y^{^{''}} + 2 y^{^{'}} + 37 y = 0, y (0) = - 3, y^{^{'}} (0) = 21

\frac{d}{d y} ((2 - y^{2} + x^{2}) e^{(1 - x^{2} - \frac{y ^{2}}{4})})

What is the tangent of f(x)=sqrt(x^2+16),\at x=-3 ?
The tangent of f(x)=sqrt(x^2+16),\at x=-3 is y=-3/5 x+16/5