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

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

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

tangent of

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

y = 2 x + 5

Solution steps

Find the slope of

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

Compute the slope of

f (x) = 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

7 x sin (x)

\int_{1}^{3} 2 x + 3 d x

\frac{d}{d x} (\frac{1}{9 x + 1})

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

\frac{d}{d x} (- 2 x + \frac{1}{x})

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