What is the tangent of f(x)=x^2-8,\at x=7 ?

The tangent of f(x)=x^2-8,\at x=7 is y=14x-57

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tangent of f(x)=x^2-8,\at x=7

tangent of

f (x) = x^{2} - 8, at x = 7

y = 14 x - 57

Solution steps

Find the tangent point:

(7, 41)

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 = 7 : m = 14

Find the line with slope m=

14

and passing through

(7, 41) : y = 14 x - 57

y = 14 x - 57

\frac{d}{d x} (8 x + 6)

f (x) = - 2 x + 5

\int \frac{1}{y ^{2} - 1} d y

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

\int 1 + (sinh (x))^{2} d x

What is the tangent of f(x)=x^2-8,\at x=7 ?
The tangent of f(x)=x^2-8,\at x=7 is y=14x-57