The curve y(x) = ax^{3} + bx^{2} + cx + 5 tou | vedclass

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(A) Given $y(x) = ax^{3} + bx^{2} + cx + 5$. Since the curve passes through $(-2, 0)$,we have $y(-2) = a(-8) + b(4) - 2c + 5 = 0$,which simplifies to

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$\frac{27}{4}$

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(A) Given $y(x) = ax^{3} + bx^{2} + cx + 5$. Since the curve passes through $(-2, 0)$,we have $y(-2) = a(-8) + b(4) - 2c + 5 = 0$,which simplifies to $-8a + 4b - 2c = -5$,or $8a - 4b + 2c = 5$ (Equation $1$). Since the curve touches the $x$-axis at $(-2, 0)$,the slope $y'(-2) = 0$. $y'(x) = 3ax^{2} + 2bx + c$. $y'(-2) = 3a(4) + 2b(-2) + c = 12a - 4b + c = 0$ (Equation $2$). Given $y'(0) = 3$,we substitute $x=0$ into $y'(x)$ to get $c = 3$ (Equation $3$). Substituting $c=3$ into Equations $1$ and $2$: $8a - 4b + 6 = 5 \implies 8a - 4b = -1$. $12a - 4b + 3 = 0 \implies 12a - 4b = -3$. Subtracting the first from the second: $(12a - 4b) - (8a - 4b) = -3 - (-1) \implies 4a = -2 \implies a = -\frac{1}{2}$. Substituting $a = -\frac{1}{2}$ into $12a - 4b = -3$: $12(-\frac{1}{2}) - 4b = -3 \implies -6 - 4b = -3 \implies -4b = 3 \implies b = -\frac{3}{4}$. Thus,$y(x) = -\frac{1}{2}x^{3} - \frac{3}{4}x^{2} + 3x + 5$. $y'(x) = -\frac{3}{2}x^{2} - \frac{3}{2}x + 3$. Setting $y'(x) = 0$: $-\frac{3}{2}(x^{2} + x - 2) = 0 \implies (x+2)(x-1) = 0$. Critical points are $x = -2$ and $x = 1$. $y''(x) = -3x - \frac{3}{2}$. At $x = 1$,$y''(1) = -3 - 1.5 = -4.5 $y(1) = -\frac{1}{2}(1)^{3} - \frac{3}{4}(1)^{2} + 3(1) + 5 = -0.5 - 0.75 + 3 + 5 = 6.75 = \frac{27}{4}$.

The curve $y(x) = ax^{3} + bx^{2} + cx + 5$ touches the $x$-axis at the point $P(-2, 0)$ and cuts the $y$-axis at the point $Q$,where the derivative $y'(0) = 3$. Find the local maximum value of $y(x)$.

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