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

The tangent of f(x)=x^2-3,(2,1) is y=4x-7

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

tangent of

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

y = 4 x - 7

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 4

Find the line with slope m=

4

and passing through

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

y = 4 x - 7

n = 0 \sum \infty \frac{n + 1}{3}

f (x) = 2 sin (x)

e^{- 2 t} (t^{2} + 4 t + 5)

\int_{a}^{2 a} (a - \frac{a ^{3}}{x ^{2}}) d x

\frac{d}{d x} (\frac{2}{x ^{2} + 5})

What is the tangent of f(x)=x^2-3,(2,1) ?
The tangent of f(x)=x^2-3,(2,1) is y=4x-7