How do you graph y = 1/2(x-3)(x+1)?

1 Answer
Apr 9, 2018

The plot is a concave upward parabola with a vertex at (1, -2) and roots at $x = 3$, and $x = - 1$.

Explanation:

$y = \frac{1}{2} \left(x - 3\right) \left(x + 1\right)$

Is a quadratic equation, so we know that the plot will be a parabola.

The equation as given is in factored form. If the factored form is written with all real numbers, it gives us the x-intercepts (roots) for the plot. If an equation is in factored form,

$y = a \left(x - {r}_{1}\right) \left(x - {r}_{2}\right)$

then, the x-intercepts for the graph are at $\left(0 , {r}_{1}\right)$, and $\left(0 , {r}_{2}\right)$. If $a$ is positive, then the plot is concave up. If $a$ is negative, then the plot is concave down.

Here, $a = \frac{1}{2}$ is positive, so we must draw our parabola concave up. Also, ${r}_{1} = 3$ and ${r}_{2} = - 1$ so the x-intercepts for the plot are $\left(0 , 3\right)$, and $\left(0 , - 1\right)$.

Finally, before we start to plot the parabola, it would be helpful to know the coordinates of the vertex of the parabola. Here's a trick! The $x$-coordinate of the vertex of a parabola is the arithmetic average of the x-intercepts (roots) for the parabola.

$x$-coordinate of the vertex = $\frac{{r}_{1} + {r}_{2}}{2}$

Here, the $x$-coordinate of the vertex = $\frac{3 - 1}{2} = 1$. We can find the $y$-coordinate of the vertex by substituting this $x$ value into our original equation.

$y$ coordinate of vertex $= \frac{1}{2} \left(1 - 3\right) \left(1 + 1\right) = - 2$

The coordinates for the vertex of this parabola are at (1, -2).

Now we are ready to plot this parabola.

graph{1/2(x-3)(x+1) [-5, 5, -5, 5]}