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

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

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

tangent of

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

y = 2 a x - a^{2} - 4

Solution steps

Compute the tangent line to the general point

x = a

Find the tangent point:

(a, a^{2} - 4)

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 = a : m = 2 a

Find the line with slope m=

2 a

and passing through

(a, a^{2} - 4) : y = 2 a x - a^{2} - 4

y = 2 a x - a^{2} - 4

\int_{5}^{9} (9 - y + 2)^{2} d y

x \to 2 lim (\frac{1}{2} x + c)

\frac{d}{d x} (\frac{x + 2}{3 x - 4})

n = 1 \sum \infty 8 \cdot (\frac{1}{5})^{n}

\int \frac{7 x + 1}{x ( x ^{2} - 1 )} d x

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