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

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Answer 1 · 2017-06-20T06:26:54

This is an upright parabola with vertex $(1, -4)$ , $x$ -intercepts $(-1, 0)$ , $(3, 0)$ and $y$ intercept $(0, -3)$

Given:

$y = (x+1)(x-3)$

Since the coefficient of $x^2$ is positive, this will be an upright (U-shaped) parabola.

When $x=0$ we find:

$y = (color(blue)(0)+1)(color(blue)(0)-3) = -3$

So the $y$ -intercept is at $(0, -3)$

Notice that $y=0$ when $x=-1$ or $x=3$

So this function has $x$ -intercepts $(-1, 0)$ and $(3, 0)$

The parabola will be symmetrical about its axis, which will be a vertical line half way between these two $x$ -intercepts, that is:

$x = 1/2((color(blue)(-1))+color(blue)(3)) = 1$

The vertex lies on the axis, so we can find its $y$ coordinate by substituting the value $x=1$ into the formula:

$y = (color(blue)(1)+1))(color(blue)(1)-3) = 2*(-2) = -4$

So the vertex is at $(1, -4)$

That's probably enough points and features to graph looking roughly like this (with axis and points indicated):

graph{(y-(x+1)(x-3))(x-1+0.00001y)((x+1)^2+y^2-0.01)(x^2+(y+3)^2-0.01)((x-1)^2+(y+4)^2-0.01)((x-3)^2+y^2-0.01) = 0 [-9.595, 10.405, -5.36, 4.64]}

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