A straight line joins the points $(-1,1)$ and $(3,9)$. Its $x$-intercept is:
$\textbf{(A)}\ -\frac{3}{2}\qquad  \textbf{(B)}\ -\frac{2}{3}\qquad  \textbf{(C)}\ \frac{2}{5}\qquad  \textbf{(D)}\ 2\qquad  \textbf{(E)}\ 3$
The slope of the line is $\frac{ \Delta y }{ \Delta x} = \frac{9-1}{3-(-1)} = 2$. Using the formula for the point-slope form of a line, we have $y-y_1 = m(x-x_1)$, so $y-1=2(x-(-1)) \to y-1=2(x+1)$.
The x-intercept is the x-value when $y=0$, so we substitute 0 for y:
\[0-1=2x+2\]
\[-1=2x+2\]
\[2x = -3\]
\[x = -\frac{3}{2} \to \boxed{\text{(A)}}\]