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

The tangent of y=2x^2-sqrt(x),(1,1) is y= 7/2 x-5/2

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

tangent of

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

y = \frac{7}{2} x - \frac{5}{2}

Solution steps

Find the slope of

y = 2 x^{2} - x : \frac{d y}{d x} = 4 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{7}{2}

Find the line with slope m=

\frac{7}{2}

and passing through

(1, 1) : y = \frac{7}{2} x - \frac{5}{2}

y = \frac{7}{2} x - \frac{5}{2}

What is the tangent of y=2x^2-sqrt(x),(1,1) ?
The tangent of y=2x^2-sqrt(x),(1,1) is y= 7/2 x-5/2