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

The tangent of f(x)=x^2-7,(2,-3) is y=4x-11

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tangent of f(x)=x^2-7,(2,-3)

tangent of

f (x) = x^{2} - 7, (2, - 3)

y = 4 x - 11

Solution steps

Find the slope of

f (x) = x^{2} - 7 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 7

(2, - 3) : m = 4

Find the line with slope m=

4

and passing through

(2, - 3) : y = 4 x - 11

y = 4 x - 11

\frac{d}{d x} ((ln (x + 1))^{2})

y = 9 x, y = x^{2} - 10, y = 0

y^{^{'}} + 3 x^{2} y = 0

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

What is the tangent of f(x)=x^2-7,(2,-3) ?
The tangent of f(x)=x^2-7,(2,-3) is y=4x-11