What is the tangent of f(x)=sqrt(x^2-10x+4),\at x=0 ?

The tangent of f(x)=sqrt(x^2-10x+4),\at x=0 is y=-5/2 x+2

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tangent of f(x)=sqrt(x^2-10x+4),\at x=0

tangent of

f (x) = x^{2} - 10 x + 4, at x = 0

y = - \frac{5}{2} x + 2

Solution steps

Find the tangent point:

(0, 2)

Find the slope of

f (x) = x^{2} - 10 x + 4 : \frac{df ( x )}{d x} = \frac{x - 5}{x ^{2} - 10 x + 4}

Compute the slope of

f (x) = x^{2} - 10 x + 4

x = 0 : m = - \frac{5}{2}

Find the line with slope m=

- \frac{5}{2}

and passing through

(0, 2) : y = - \frac{5}{2} x + 2

y = - \frac{5}{2} x + 2

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What is the tangent of f(x)=sqrt(x^2-10x+4),\at x=0 ?
The tangent of f(x)=sqrt(x^2-10x+4),\at x=0 is y=-5/2 x+2