finding an equation from a graph – Online math tutoring

Question

This post about how to find a quadratic function’s equation from its graph is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

This graph represents a polynomial function $y=g(x)$ . Which of the following could be g‘s equation?

$g(x)=x^2-2x-3$
$g(x)=-x^2+2x+3$
$g(x)=(x-1)(x+3)$
$g(x)=-(x-1)^2-4$

Solution

What can you tell from the graph?

You can read the coordinates of the x-intercepts, the y-intercept, and the turning point.

	x-intercepts		(-1, 0), (3, 0)
	y-intercept		(0, -3)
	turning point		(1, -4)

The x-intercepts are clues to g‘s factors: The x-intercept (-1, 0) tells you that $x+1$ is a zero, and the x-intercept (3, 0) tells you that $x-3$ is a factor.

Sidebar: The factor theorem says that a polynomial that has a polynomial has a factor $x-c$ has a root at c, and a polynomial that has a root at c has a factor $x-c$ . A root is the x-value at the x-intercept: for this example, -1 and 3. So the factors are $x+1$ and $x-3$ .

Multiply those factors together and see what you get:

${\color{DarkBlue} (x+1)(x-3)=x^2-2x-3}$

The graph looks like a parabola, a second-degree curve. And $x^2-2x-3$ is second degree. So you may be done. How can you check? Use one of the two other points you can read from the graph. The y-intercept is (0, -3). Plug in 0 for x and see if the equation gives you -3, the y-intercept.

${\color{DarkBlue} g(x)=x^2-2x-3}$

then

${\color{DarkBlue} g(0)=0^2-2 \cdot 0-3}$

And as we saw from the graph, the y-intercept is (0, -3). Check. So answer choice #1 is the correct one.

That is one way to find a quadratic function’s equation from its graph.

Alternatively, since this question is multiple choice, you could try each answer choice. The easiest way to do that, I think, is to substitute 0 for x and see if you get that $\inline \fn_phv g\left ( 0 \right )=-3$ , the y-intercept:

$g(x)=x^2-2x-3$ . Then $g(0)=-3$ . This is the answer we found. It works.
$g(x)=-x^2+2x+3$ . $g(0)=3\neq -3$ . Not this one.
$g(x)=(x-1)(x+3)$ . $g(0)=-3$ . Looks like this works.
$g(x)=-(x-1)^2-4$ . $g(0)=-5$ . Does not work.

So we have two contenders, answer choice #1 and answer choice #3. That means we have to try at least one more point besides the y-intercept. We have already done that for answer choice #1; we know the x-intercepts work for it. Let’s see if $y=0$ at $x=-1$ for answer choice #3, $g(x)=(x-1)(x+3)$ .