What is the tangent of f(x)=x^2-5,(2,-1) ?

The tangent of f(x)=x^2-5,(2,-1) is y=4x-9

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tangent of f(x)=x^2-5,(2,-1)

tangent of

f (x) = x^{2} - 5, (2, - 1)

y = 4 x - 9

Solution steps

Find the slope of

f (x) = x^{2} - 5 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 5

(2, - 1) : m = 4

Find the line with slope m=

4

and passing through

(2, - 1) : y = 4 x - 9

y = 4 x - 9

\frac{( 5 x - 4 )}{x - 1}

\frac{d}{d x} (- s t)

\int \frac{1}{x ^{2} x} d x

3 ln (x)

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

What is the tangent of f(x)=x^2-5,(2,-1) ?
The tangent of f(x)=x^2-5,(2,-1) is y=4x-9