At what points of the curve y = frac{2}{3}x^3 | vedclass

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(A) Given the curve $y = \frac{2}{3}x^3 + \frac{1}{2}x^2$. Differentiating with respect to $x$,we get the slope of the tangent: $\frac{dy}{dx} = 2x^2

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$\left( \frac{1}{2}, \frac{5}{24} \right)$ and $\left( -1, -\frac{1}{6} \right)$

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(A) Given the curve $y = \frac{2}{3}x^3 + \frac{1}{2}x^2$. Differentiating with respect to $x$,we get the slope of the tangent: $\frac{dy}{dx} = 2x^2 + x$ ... $(i)$ Since the tangent makes an equal angle with the axes,the angle of inclination $	heta$ must be $45^\circ$ or $135^\circ$ (i.e.,$\pm 45^\circ$). Thus,the slope $m = 	an(\pm 45^\circ) = \pm 1$. Case $1$: $\frac{dy}{dx} = 1$ $2x^2 + x = 1 \implies 2x^2 + x - 1 = 0$ $(2x - 1)(x + 1) = 0 \implies x = \frac{1}{2}$ or $x = -1$. Case $2$: $\frac{dy}{dx} = -1$ $2x^2 + x = -1 \implies 2x^2 + x + 1 = 0$. The discriminant $D = 1^2 - 4(2)(1) = 1 - 8 = -7 Now,find the corresponding $y$-coordinates: For $x = \frac{1}{2}$,$y = \frac{2}{3}(\frac{1}{8}) + \frac{1}{2}(\frac{1}{4}) = \frac{1}{12} + \frac{1}{8} = \frac{2+3}{24} = \frac{5}{24}$. For $x = -1$,$y = \frac{2}{3}(-1)^3 + \frac{1}{2}(-1)^2 = -\frac{2}{3} + \frac{1}{2} = \frac{-4+3}{6} = -\frac{1}{6}$. Therefore,the required points are $\left( \frac{1}{2}, \frac{5}{24} ight)$ and $\left( -1, -\frac{1}{6} ight)$.

At what points of the curve $y = \frac{2}{3}x^3 + \frac{1}{2}x^2$ does the tangent make an equal angle with the axes?

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