图中显示了曲线 $y = \sqrt{3x + 4}$ 的一部分。直线 $AB$ 在点 $A$ 处与曲线相切。点 $B$ 和 $C$ 位于 $x$ 轴上，且 $C$ 是曲线上的一个点。已知该曲线在点 $A$ 处的切线斜率为 $\frac{3}{2}$。求

1. Find the derivative of the curve equation.

$y = (3x + 4)^{\frac{1}{2}}$
$\frac{dy}{dx} = \frac{1}{2}(3x + 4)^{-\frac{1}{2}} \cdot 3 = \frac{3}{2\sqrt{3x + 4}}$

2. Equate the derivative to the given gradient to find the $x$-coordinate of $A$.

Given gradient is $\frac{3}{2}$:
$\frac{3}{2\sqrt{3x + 4}} = \frac{3}{2}$
$\sqrt{3x + 4} = 1$
$3x + 4 = 1$
$3x = -3$
$x = -1$

3. Find the $y$-coordinate of $A$.

Substitute $x = -1$ into the curve equation:
$y = \sqrt{3(-1) + 4} = \sqrt{1} = 1$
Coordinates of $A$ are $(-1, 1)$.

4. Find the equation of the tangent line $AB$.

Using the point $(-1, 1)$ and gradient $m = \frac{3}{2}$:
$y - 1 = \frac{3}{2}(x - (-1))$
$y - 1 = \frac{3}{2}(x + 1)$
$y = \frac{3}{2}x + \frac{3}{2} + 1$
$y = \frac{3}{2}x + \frac{5}{2}$

1. Concisely find the equation of the tangent line $AB$ (required from part i).

$\frac{dy}{dx} = \frac{3}{2\sqrt{3x+4}}$
Given gradient is $\frac{3}{2}$:
$\frac{3}{2\sqrt{3x+4}} = \frac{3}{2} \implies \sqrt{3x+4} = 1 \implies 3x+4 = 1 \implies x = -1$
When $x = -1$, $y = \sqrt{3(-1)+4} = 1$. So $A$ is $(-1, 1)$.
Equation of $AB$: $y - 1 = \frac{3}{2}(x - (-1)) \implies y = \frac{3}{2}x + \frac{5}{2}$

2. Find the coordinates of $B$ by finding the $x$-intercept of the line $AB$.

At $B$, $y = 0$:
$0 = \frac{3}{2}x + \frac{5}{2}$
$\frac{3}{2}x = -\frac{5}{2} \implies x = -\frac{5}{3}$
Coordinates of $B$ are $\left(-\frac{5}{3}, 0\right)$.

3. Find the coordinates of $C$ by finding the $x$-intercept of the curve.

At $C$, $y = 0$:
$0 = \sqrt{3x+4}$
$3x + 4 = 0 \implies x = -\frac{4}{3}$
Coordinates of $C$ are $\left(-\frac{4}{3}, 0\right)$.