What is the tangent of y=-(x^2)/(40),(-4sqrt(5),-2) ?

The tangent of y=-(x^2)/(40),(-4sqrt(5),-2) is y= 1/(sqrt(5))x+2

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tangent of y=-(x^2)/(40),(-4sqrt(5),-2)

tangent of

y = - \frac{x ^{2}}{40}, (- 45, - 2)

y = \frac{1}{5} x + 2

Solution steps

Find the slope of

y = - \frac{x ^{2}}{40} : \frac{d y}{d x} = - \frac{x}{20}

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{1}{5}

Find the line with slope m=

\frac{1}{5}

and passing through

(- 45, - 2) : y = \frac{1}{5} x + 2

y = \frac{1}{5} x + 2

\frac{d}{d x} (f (x) (2 x^{2} - 4))

4 x, x^{3}

\frac{\partial}{\partial x} ((x + 1)^{2})

5 + 5 t^{2} - t - t^{3}

y \to - 1 lim (\frac{y + 1}{y ^{3} + 1})

What is the tangent of y=-(x^2)/(40),(-4sqrt(5),-2) ?
The tangent of y=-(x^2)/(40),(-4sqrt(5),-2) is y= 1/(sqrt(5))x+2