What is the tangent of y=x^2-sqrt(x),(4,14) ?

The tangent of y=x^2-sqrt(x),(4,14) is y= 31/4 x-17

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tangent of y=x^2-sqrt(x),(4,14)

tangent of

y = x^{2} - x, (4, 14)

y = \frac{31}{4} x - 17

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{31}{4}

Find the line with slope m=

\frac{31}{4}

and passing through

(4, 14) : y = \frac{31}{4} x - 17

y = \frac{31}{4} x - 17

\int_{0}^{3} 2 π (3 - x) (9 - x^{2}) d x

- 4 e^{- x} + 1

\int 8 x e^{x} d x

f (x) = (1 + 2 x)^{2}, at x = 3

g

What is the tangent of y=x^2-sqrt(x),(4,14) ?
The tangent of y=x^2-sqrt(x),(4,14) is y= 31/4 x-17