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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.

y=g(x). This parabola has x-intercepts at x=-1 and x=3, y-intercept at y=-3, and vertex at (0, -4).

This graph represents a polynomial function . Which of the following could be g‘s equation?


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 is a zero, and the x-intercept (3, 0) tells you that is a factor.

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

Multiply those factors together and see what you get:

The graph looks like a parabola, a second-degree curve. And 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

then

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 \fn_phv g\left ( 0 \right )=-3, the y-intercept:

  1. .  Then . This is the answer we found. It works.
  2. . Not this one.
  3. .  Looks like this works.
  4. . 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  at  for answer choice #3, .

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

Related: “Finding a Second Degree Equation from Its Solutions.”

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.

Completing the Square

Completing the Square

Problem

Express the equation y=3x^2-x+1  in the form y=a(x-h)^2+k,  where a,  h,  and k are constants.

Solution

You’re asked to convert a quadratic equation into vertex format. The original form — y=3x^2-x+1 — makes the y-intercept easy to see: It’s (0, 1). y=a(x-h)^2+k is called vertex format; it makes the vertex’s coordinate easy to see: It’s (h, k).

How do you convert it?

By a method called “completing the square.” It’s called that (I think) because you start with something that does not obviously contain a squared binomial and you end with something that does.

These are the steps:

{\color{Blue} 3x^2-x+1}
{\color{Blue} 3x^2-x} Ditch the last term. We’ll grab it back later, but now we are working with just the first two terms.
{\color{Blue} 3(x^2-x/3)} Factor out the number that multiplies the first term. Yes, in this case that leaves you with a fraction in the second term. It’s just something you have to deal with.
{\color{Blue} -1/3} Find the multiplier of x in the second term.
{\color{Blue} -1/6} Divide that number by 2.
{\color{Blue} 1/36} Square the result
{\color{Blue} x^2-x/3+1/36} Add that result to x^2-x/3  from 3(x^2-x/3)  above.
{\color{Blue} (x-1/6)(x-1/6)}

{\color{Blue} =(x-1/6)^2}

That should be a perfect square binomial. Express it that way.
 

{\color{Blue} 3*(1/36)=1/12}

Multiply (x-1/6)^2  by 3. That’s the 3 we factored out – it’s still there; we just stopped writing it while we focused on x^2-x/3. When we added 1/36  to x^2-x/3, we effectively added 3*1/36=1/12  to  3(x^2-x/3).
{\color{Blue} 3x^2-x+1}

{\color{Blue} =3(x-1/6)^2+1-1/12}

{\color{Blue} =3(x-1/6)^2+11/12}

We started with y=3x^2-x+1 and added 1/12 to create a perfect square. In adding 1/12, we changed the value of the expression we were given. To change it back to what it was, we need to subtract 1/12.

That’s the answer: {\bold{\color{Blue} 3(x-1/6)^2+11/12}}.

Why this works

Consider a perfect square binomial that looks like this: (x+d)^2=x^2+2dx+d^2. By “like this” I mean starting with x, not with a constant. Note that the multiplier of the second term (that’s 2d), divided by 2 (that’s d) and squared (that’s d^2), is the third term. That means that a trinomial that starts with x, not a constant, if it’s not a perfect square – that is, if it can’t be expressed as (x+d)^2 – can be made into a perfect square just by the addition or subtraction of a constant equal to the second term’s multiplier divided by 2 and squared.

For example, you can make x^2-6x+3  into a square binomial – something that can be expressed as (x+d)^2 – by adding or subtracting some number. What number? Here is how you find it:

  • Just for now, ignore that last term. Work with x^2-6x.
  • The second term is -6x. The multiplier of x is -6.
  • Divide that number, -6, by 2. That’s -3.
  • Square the result: 9.
  • Add that 9 to x^2-6x. That gives you x^2-6x+9, which should be a perfect square trinomial.
  • Rewrite that in square form: (x-3)^2.
  • By ignoring the last term, you subtract 3. After that you add 9. In total you add -3+9=6. You need something that’s equal to what you started with, so subtract 6 from the perfect square you got: x^2-6x+3 = (x-3)^2-6.

There. That’s how you compile a square.

Oh, Jill, you’re making it too simple. I need to be able to do problems where a number is multiplying x^2.

Yes, that’s right. But you can factor that number out so that you can still solve the problem as if it started with x^2.

3x^2-x+1=3(x^2-x/3) plus something we are temporarily ignoring.

And this is what we did above, where we converted 3x^2-x+1 into 3(x-1/6)^2+11/12.