What is the tangent of f(x)=sqrt(-x^2+25),\at x=-4 ?

The tangent of f(x)=sqrt(-x^2+25),\at x=-4 is y= 4/3 x+25/3

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tangent of f(x)=sqrt(-x^2+25),\at x=-4

tangent of

f (x) = - x^{2} + 25, at x = - 4

y = \frac{4}{3} x + \frac{25}{3}

Solution steps

Find the tangent point:

(- 4, 3)

Find the slope of

f (x) = - x^{2} + 25 : \frac{df ( x )}{d x} = - \frac{x}{- x ^{2} + 25}

Compute the slope of

f (x) = - x^{2} + 25

x = - 4 : m = \frac{4}{3}

Find the line with slope m=

\frac{4}{3}

and passing through

(- 4, 3) : y = \frac{4}{3} x + \frac{25}{3}

y = \frac{4}{3} x + \frac{25}{3}

\frac{10 x + 1}{x + 100}

7^{x^{7}}

x \to - 1 lim (- x^{2} + 3)

\frac{d y}{d t} - 3 y = t e^{2 t}

\int \frac{x}{( a ^{2} + x ^{2} ) ^{\frac{3}{2}}} d x

What is the tangent of f(x)=sqrt(-x^2+25),\at x=-4 ?
The tangent of f(x)=sqrt(-x^2+25),\at x=-4 is y= 4/3 x+25/3