What is the tangent of f(x)=x^2-8,\at x=4 ?

The tangent of f(x)=x^2-8,\at x=4 is y=8x-24

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tangent of f(x)=x^2-8,\at x=4

tangent of

f (x) = x^{2} - 8, at x = 4

y = 8 x - 24

Solution steps

Find the tangent point:

(4, 8)

Find the slope of

f (x) = x^{2} - 8 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 8

x = 4 : m = 8

Find the line with slope m=

8

and passing through

(4, 8) : y = 8 x - 24

y = 8 x - 24

x y^{^{'}} = 3 y - 6 x^{2}

\int cos (π x) cos (4 π x) d x

3 x, 3, \frac{3}{x}

\int \frac{x}{2 ln ( x )} d x

x \to 1 lim (x + 2)

What is the tangent of f(x)=x^2-8,\at x=4 ?
The tangent of f(x)=x^2-8,\at x=4 is y=8x-24