The curve $y=ax^3+bx^2+cx+5$ touches the X axis at the point $\left(-2,0\right)$ and cuts the Y axis at a point where its gradient is 3. Find a, b and c.
When the curve $y=ax^3+bx^2+cx+5$ cuts the Y axis, the X coordinate of the point is 0.
$\implies$ $y=5$
$\therefore$ $\;\;$ At the point $\left(0,5\right)$, the gradient of the curve is 3
Differentiating the curve $y=ax^3+bx^2+cx+5$ with respect to x gives
$\dfrac{dy}{dx}=3ax^2+2bx+c \;\; \cdots$ (1)
$\therefore$ $\;\;$ $\dfrac{dy}{dx}\bigg|_{\left(0,5\right)}=c$
The gradient of the curve at $\left(0,5\right)$ is 3
$\implies$ $c=3 \;\; \cdots$ (2)
The curve $y=ax^3+bx^2+cx+5$ touches the X axis at the point $\left(-2,0\right)$
$\implies$ $0=\left(-2\right)^3 a + \left(-2\right)^2 b -2c+5$
i.e. $8a-4b+2c = 5$
Substituting the value of c gives
$8a-4b=-1 \;\; \cdots$ (3)
Since the curve touches the X axis at $\left(-2,0\right)$
$\implies$ X axis is tangent to the curve at $\left(-2,0\right)$
$\implies$ $\dfrac{dy}{dx}\bigg|_{\left(-2,0\right)}=0$
$\therefore$ $\;\;$ We have from equations (1) and (2)
$3 \times \left(-2\right)^2 a + 2 \times \left(-2\right)b + 3 = 0$
i.e. $12a-4b=-3 \;\; \cdots$ (4)
Solving equations (3) and (4) simultaneously gives
$4a = -2$ $\implies$ $a = -\dfrac{1}{2}$
Substituting the value of a in equation (3) gives
$8 \times \left(\dfrac{-1}{2}\right)-4b=-1$
i.e. $4b = -3$ $\implies$ $b = -\dfrac{3}{4}$
$\therefore$ $\;\;$ $a=-\dfrac{1}{2}$, $b = -\dfrac{3}{4}$, $c=3$