What is the tangent of f(x)=sqrt(x^2+9),\at x=-4 ?

The tangent of f(x)=sqrt(x^2+9),\at x=-4 is y=-4/5 x+9/5

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tangent of f(x)=sqrt(x^2+9),\at x=-4

tangent of

f (x) = x^{2} + 9, at x = - 4

y = - \frac{4}{5} x + \frac{9}{5}

Solution steps

Find the tangent point:

(- 4, 5)

Find the slope of

f (x) = x^{2} + 9 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 9}

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{4}{5}

Find the line with slope m=

- \frac{4}{5}

and passing through

(- 4, 5) : y = - \frac{4}{5} x + \frac{9}{5}

y = - \frac{4}{5} x + \frac{9}{5}

\int x os (x) d x

n = 1 \sum \infty \frac{e ^{n}}{n}

t y^{^{'}} + 2 y = t^{2} - t + 1, y (1) = 0.5

d er i v a t i v e sin^{2} (14 x)

t an g e n t f (x) = 2 x^{2} - 4, at x = 8

What is the tangent of f(x)=sqrt(x^2+9),\at x=-4 ?
The tangent of f(x)=sqrt(x^2+9),\at x=-4 is y=-4/5 x+9/5