Question

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

Answer 1

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

Explanation:

`y=1/2(x-3)(x+1)`

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(x-r_1)(x-r_2)`

then, the x-intercepts for the graph are at `(0, r_1)`, and `(0, r_2)`. If `a` is positive, then the plot is concave up. If `a` is negative, then the plot is concave down.

Here, `a=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 `(0, 3)`, and `(0,-1)`.

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 = `(r_1+r_2)/2`

Here, the `x`-coordinate of the vertex = `(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 `=1/2(1-3)(1+1)=-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]}