What is the tangent of f(x)=sqrt(x^2-x+10),\at x=3 ?

The tangent of f(x)=sqrt(x^2-x+10),\at x=3 is y= 5/8 x+17/8

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tangent of f(x)=sqrt(x^2-x+10),\at x=3

tangent

y = \frac{5}{8} x + \frac{17}{8}

Solution steps

Find the tangent point:

(3, 4)

Find the slope of

f (x) = x^{2} - x + 10 : \frac{df ( x )}{d x} = \frac{2 x - 1}{2 x ^{2} - x + 10}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{5}{8}

Find the line with slope m=

\frac{5}{8}

and passing through

(3, 4) : y = \frac{5}{8} x + \frac{17}{8}

y = \frac{5}{8} x + \frac{17}{8}

What is the tangent of f(x)=sqrt(x^2-x+10),\at x=3 ?
The tangent of f(x)=sqrt(x^2-x+10),\at x=3 is y= 5/8 x+17/8