What is the tangent of y=4-x^2,(-1,3) ?

The tangent of y=4-x^2,(-1,3) is y=2x+5

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tangent of y=4-x^2,(-1,3)

tangent of

y = 4 - x^{2}, (- 1, 3)

y = 2 x + 5

Solution steps

Find the slope of

y = 4 - x^{2} : \frac{d y}{d x} = - 2 x

Compute the slope of

y = 4 - x^{2}

(- 1, 3) : m = 2

Find the line with slope m=

2

and passing through

(- 1, 3) : y = 2 x + 5

y = 2 x + 5

\frac{1}{3 s + 1}

y^{^{''}} + 14 y^{^{'}} + 49 y = t^{- 2} e^{- 7 t}

\int (z + 2)^{2} d z

\frac{d}{d x} (cos (2^{x} \cdot x^{5}))

\frac{d}{d x} (9 + 9 - x)