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-5

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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 - 5

Solution steps

Find the tangent point:

(1, - 3)

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, - 3) : y = 2 x - 5

y = 2 x - 5

(y^{2} + 9) d x - \frac{1}{( 2 x + 1 ) ^{2}} d y = 0

(\frac{x + 3}{x - 3})^{5}

\int 5 x 1 - x^{2} d x

\frac{d}{d x} (\frac{3 x + 2}{x + 1})

\int_{0}^{1} 8^{x} d 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-5