[tex]\bf f(x)=\cfrac{3}{2}x+1\implies \cfrac{df}{dx}=\cfrac{3}{2}\qquad \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}
\\\\\\
\stackrel{(-2,-2)}{y-(-2)=\cfrac{3}{2}[x-(-2)]}\implies y+2=\cfrac{3}{2}(x+2)\implies y+2=\cfrac{3}{2}x+3
\\\\\\
y=\cfrac{3}{2}x+1[/tex]
you might notice the equation kinda went for a full circle runaround, ended at square one, well, that's because, the equation is linear and therefore a line and the tangent line at such point, will be the same as the line's, and it's equation will also be the same.
