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

The tangent of y=sqrt(x^2-x+2),\at x=2 is y= 3/4 x+1/2

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tangent of y=sqrt(x^2-x+2),\at x=2

tangent of

y = x^{2} - x + 2, at x = 2

y = \frac{3}{4} x + \frac{1}{2}

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

y = x^{2} - x + 2 : \frac{d y}{d x} = \frac{2 x - 1}{2 x ^{2} - x + 2}

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{3}{4}

Find the line with slope m=

\frac{3}{4}

and passing through

(2, 2) : y = \frac{3}{4} x + \frac{1}{2}

y = \frac{3}{4} x + \frac{1}{2}

\frac{d}{d x} ((x + 3)^{2})

x \to 0 lim (x sin (\frac{8}{x}))

(6 + x) \frac{d y}{d x} = 3 y

\frac{d}{d x} (\frac{4}{x + 2})

\frac{\partial}{\partial x} (4 x^{2} + y^{2})

What is the tangent of y=sqrt(x^2-x+2),\at x=2 ?
The tangent of y=sqrt(x^2-x+2),\at x=2 is y= 3/4 x+1/2