y1 = 2x√x
y2 = 4 + 6x
y=
The general equation of a line is $\displaystyle y - y_1 = m(x - x_1)$ where m is the gradient and $\displaystyle (x_1, y_1)$ is a known point on the line.
In your question, parallel to y2 = 4 + 6x means that m = 6.
To get a known point:
x-coordinate: Solve $\displaystyle \frac{dy_1}{dx} = 6$ for x.
y-coordinate: Substitute the value of x found above into $\displaystyle y_1 = 2 x \sqrt{x} = 2 x^{3/2}$.
x-coordinate: Solve $\displaystyle \frac{dy_1}{dx} = 6$ for x:
$\displaystyle \frac{dy_1}{dx} = \frac{3}{2} \times 2x^{\frac{3}{2} - 1} = 3 x^{1/2}$.
So solve $\displaystyle 3 x^{1/2} = 6$:
$\displaystyle \Rightarrow x^{1/2} = 2$
$\displaystyle \Rightarrow x = 4$.
Now substitute x = 4 into $\displaystyle y_1 = 2 x^{3/2} = 2 x x^{1/2} = 2 x \sqrt{x}$:
$\displaystyle y = 2 (4) \sqrt{4} = 2 (4) (2) = 16$.
So a known point is (4, 16) ........
So $\displaystyle x_1 = 4$ and $\displaystyle y_1 = 16$ .......