What is the tangent of x^2+y^2=25,(-4,3) ?

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

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tangent of x^2+y^2=25,(-4,3)

tangent of

x^{2} + y^{2} = 25, (- 4, 3)

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

Solution steps

Find the slope of

x^{2} + y^{2} = 25 : \frac{d y}{d x} = - \frac{x}{y}

Compute the slope of

x^{2} + y^{2} = 25

(- 4, 3) : 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}

f (x) = \frac{4 x}{x ^{2} + 3}, at x = 1

x \to 0 + lim (cos (\frac{π}{x}))

\frac{d}{d x} ((1 + x)^{\frac{1}{3}})

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

y = sin (x^{3})

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