Fourier transform of the product
First of all, your graph does not match the equation you gave for the function. From the hint of msm, we should use the following property of the Fourier Transform:

$$
\mathcal{F}\left\{\frac{d}{dt}x(t)\right\}=j2\pi fX(f)
$$

with 

$$
\mathcal{F}\{x(t)\} = X(f)
$$

Now, what's the derivative of your function in the graph? Its given by

$$
x'(t)=\frac{1}{4}\text{rect}\left(\frac{t-2}{4}\right) - \text{rect}(t-\frac{9}{2})
$$

Now, let's do the Fourier Transform of the derivative:

$$
j2\pi fX(f)=\text{sinc}(4f)\exp(-j2\pi 2f) - \text{sinc}(f)\exp(-j2\pi \frac{9}{2}f)
$$

with $\text{sinc}(x)=\frac{\sin(\pi x)}{\pi x}$.

Now, finally dividing by $j2\pi f$ you get

$$
X(f) = \frac{1}{j2\pi f}\left(\text{sinc}(4f)\exp(-j2\pi2f)-\text{sinc}(f)\exp(-j2\pi\frac{9}{2}f)\right).
$$

You can try to simplify this expression, but it might not yield something that is better to understand.