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

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

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tangent of f(x)=x^2+4,\at x=-1

tangent of

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

y = - 2 x + 3

Solution steps

Find the tangent point:

(- 1, 5)

Find the slope of

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

Compute the slope of

f (x) = x^{2} + 4

x = - 1 : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x + 3

\int \frac{( x ^{2} - x + 18 )}{x ^{3} + 3 x} d x

\int (3 x^{4}) d x

f (x) = x - 3, at x = 7

2 t y^{^{'}} = 8 y

\frac{d}{d x} (8 ln (sin (x)))

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