A curve with equation of the form y = ax^4 + | vedclass

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(C) Given the equation $y = ax^4 + bx^3 + cx + d$.$1$. The gradient is zero at $(0, 1)$,so $\frac{dy}{dx} = 4ax^3 + 3bx^2 + c$. At $x=0$,$\frac{dy}{dx

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$x < -1$

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(C) Given the equation $y = ax^4 + bx^3 + cx + d$.$1$. The gradient is zero at $(0, 1)$,so $\frac{dy}{dx} = 4ax^3 + 3bx^2 + c$. At $x=0$,$\frac{dy}{dx} = c = 0$.$2$. The point $(0, 1)$ lies on the curve,so $1 = a(0)^4 + b(0)^3 + c(0) + d$,which gives $d = 1$.$3$. The curve touches the $x$-axis at $(-1, 0)$,so $0 = a(-1)^4 + b(-1)^3 + 1$,which simplifies to $a - b + 1 = 0$,or $b = a + 1$.$4$. Since it touches the $x$-axis at $(-1, 0)$,the gradient at $x = -1$ is also zero: $\frac{dy}{dx} = 4ax^3 + 3(a+1)x^2 = 0$ at $x = -1$.$5$. Substituting $x = -1$: $4a(-1)^3 + 3(a+1)(-1)^2 = -4a + 3a + 3 = 0$,which gives $a = 3$. Then $b = 3 + 1 = 4$.$6$. The equation is $y = 3x^4 + 4x^3 + 1$. The derivative is $\frac{dy}{dx} = 12x^3 + 12x^2 = 12x^2(x + 1)$.$7$. For a negative gradient,$\frac{dy}{dx} < 0$,so $12x^2(x + 1) < 0$. Since $12x^2 \ge 0$,we must have $x + 1 < 0$ and $x 
eq 0$,which implies $x < -1$.

$A$ curve with equation of the form $y = ax^4 + bx^3 + cx + d$ has zero gradient at the point $(0, 1)$ and also touches the $x$-axis at the point $(-1, 0)$. Then the values of $x$ for which the curve has a negative gradient are:

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