What is the tangent of y=x^2-sqrt(x),(1,0) ?

The tangent of y=x^2-sqrt(x),(1,0) is y= 3/2 x-3/2

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tangent of y=x^2-sqrt(x),(1,0)

tangent of

y = x^{2} - x, (1, 0)

y = \frac{3}{2} x - \frac{3}{2}

Solution steps

Find the slope of

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

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}{2}

Find the line with slope m=

\frac{3}{2}

and passing through

(1, 0) : y = \frac{3}{2} x - \frac{3}{2}

y = \frac{3}{2} x - \frac{3}{2}

What is the tangent of y=x^2-sqrt(x),(1,0) ?
The tangent of y=x^2-sqrt(x),(1,0) is y= 3/2 x-3/2