# How Can You Find a Quadratic’s Equation from Its Graph?

## 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 $\inline \fn_jvn y=g(x)$. Which of the following could be g‘s equation?

1. $\inline \fn_jvn g(x)=x^2-2x-3$
2. $\inline \fn_jvn g(x)=-x^2+2x+3$
3. $\inline \fn_jvn g(x)=(x-1)(x+3)$
4. $\inline \fn_jvn g(x)=-(x-1)^2-4$

## 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 $\fn_phv x+1$ is a zero, and the x-intercept (3, 0) tells you that $\fn_phv x-3$ is a factor.

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

Multiply those factors together and see what you get:

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

The graph looks like a parabola, a second-degree curve. And $\fn_phv x^2-2x-3$ 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.

If

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

then

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

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 and see if you get that $\inline&space;\fn_phv&space;g\left&space;(&space;0&space;\right&space;)=-3$, the y-intercept:

1. $\inline \fn_jvn g(x)=x^2-2x-3$.  Then $\inline \fn_jvn g(0)=-3$. This is the answer we found. It works.
2. $\inline \fn_jvn g(x)=-x^2+2x+3$$\inline \fn_jvn g(0)=3\neq -3$. Not this one.
3. $\inline \fn_jvn g(x)=(x-1)(x+3)$$\inline \fn_jvn g(0)=-3$.  Looks like this works.
4. $\inline \fn_jvn g(x)=-(x-1)^2-4$$\inline \fn_jvn 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 $\inline \fn_jvn y=0$ at $\inline \fn_jvn x=-1$ for answer choice #3, $\inline \fn_jvn g(x)=(x-1)(x+3)$.

$\fn_jvn {\color{DarkBlue} g(-1)=(-1-1)(-1+3)=-2 \cdot 2 = -4 \neq 0}$

So answer choice #3 does not work after all, and answer choice #1 is the correct one.

For more about the end behavior of polynomial functions, see “End Behavior, Degree, and Leading Coefficient.”

The Accuplacer sample problem on which this one is based is #10 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

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