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

The tangent of f(x)=x^2-3,\at x=2 is y=4x-7

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tangent of f(x)=x^2-3,\at x=2

tangent of

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

y = 4 x - 7

Solution steps

Find the tangent point:

(2, 1)

Find the slope of

f (x) = x^{2} - 3 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 3

x = 2 : m = 4

Find the line with slope m=

4

and passing through

(2, 1) : y = 4 x - 7

y = 4 x - 7

n = 2 \sum \infty \frac{2}{5 ^{n}}

\int_{- 6}^{- 1} 2 x^{\frac{1}{3}} d x

\int \frac{5 + 14 x}{49 + 25 x ^{2}} d x

x \to 0 lim (tan (e^{x} - 1))

y^{^{'}} - 4 y = e^{- 4 t}

What is the tangent of f(x)=x^2-3,\at x=2 ?
The tangent of f(x)=x^2-3,\at x=2 is y=4x-7