Discriminant, Roots, and Graph

[0313phd]

This is a problem that I am not sure about. It is difficult to describe because it involves graphs of parabolas that I will
describe as best I can. PROBLEM: Which of the following could be the graph of y=ax +bxsquared + c, where bsquared -4ac=0?

1. Parabola one is negative, turned downward, with vertex high on y axis, not at origin. Two x intercepts 2. Parabola is positive, turned upward, vertex on x axis only, high,upper portion and vertex does not cross y axis. 3 Vertex higher on y axis. Positive, vertex is the origin. 4. Negative, lower than origin, no x intercept. 5. Two x intercepts, positive.

I ruled out 1and 5 because they had 2 x intercepts. I ruled out 4 because it was negative and had no x intercept. I ruled out 3 because its vertex was the origin. Answer is 2. Am I thinking correctly? 0313

 

[JeffM]

This is a problem that I am not sure about. It is difficult to describe because it involves graphs of parabolas that I will
describe as best I can. PROBLEM: Which of the following could be the graph of y=ax +bxsquared + c, where bsquared -4ac=0? This is mildly confusing because the CONVENTIONAL equation for a parabola is ax[sup:3zk600pu]2[/sup:3zk600pu] + bx + c rather than bx[sup:3zk600pu]2[/sup:3zk600pu] + ax + c. The quadratic formula is usually given in terms of the conventional notation so (b[sup:3zk600pu]2[/sup:3zk600pu] - 4ac) is the conventional form of the discriminant, but it is not the discriminant of the equation you have provided. There is nothing logically necessary about using the conventional forms, but this problem SEEMS to jump about from the conventional to unconventional in an odd way. Are you sure you wrote the equation down correctly?

 

[0313phd]

Jeff- You're right. I erred, the correct equation in the problem is axsquared + bx +c, where bsquared + 4ac= 0. Sorry. 0313

 

[JeffM]

OK

y = ax[sup:nn4lkpse]2[/sup:nn4lkpse] + bx + c.

y is approximately equal to what when the absolute value of x is very very large?

 

[JeffM]

OK That prior question seemed to be no help.

Let's try a different one.

If a is positive, then ax[sup:1wtnycej]2[/sup:1wtnycej] is positive everywhere, right?

And as the absolute value of x gets larger then eventually ax[sup:1wtnycej]2[/sup:1wtnycej] > (bx + c), no matter what finite values b and c have, right?

So if a is positive, do the arms of the parabola reach upward or downward?

If a is negative, do the arms of the parabola reach upward or downward?

 

[soroban]

Hello, 0313phd!

Your descriptions are quite fuzzy.
I'll do my best and make sketches.
Also, there is an obvious typo in the equation of the parabola; I'll correct it.

\(\displaystyle \text{Which of the following could be the graph of }\:y\:=\:ax^2\,+\,bx\,+\,c,\:\text{ where } b^2\,-\,4ac \:=\:0\,?\)

  [1]       |             : [2] |                     :     [3]   |
            *             :     | *               *   :   *       |       *
        *   |   *         :     |                     :           |
      *     |     *       :     |  *             *    :    *      |      *
  ---*------+------*--    :     |   *           *     :     *     |     *
            |             :     |     *       *       :       *   |   *
    *       |       *     :  ---+---------*--------   :   --------*--------
            |             :     |                     :           |

   [4]      |             :   [5]
    --------+--------     :               |            
            *             :       *       |       *   
        *   |   *         :               |           
      *     |     *       :        *      |      *    
     *      |      *      :   ------*-----+-----*----     
            |             :           *   |   *       
    *       |       *     :               *           
            |             :               |

I ruled out 1 and 5 because they had 2 x-intercepts. . Right!

I ruled out 4 because it was negative and had no x-intercept. . Yes!

I ruled out 3 because its vertex was the origin. . no

Answer is 2.
. . Am I thinking correctly?

\(\displaystyle \text{If }b^2 - 4ac \,=\,0,\,\text{ then the parabola has }one\:x\text{-intercept.}\)
. . \(\displaystyle \text{The parabola is }tangent\text{ to the }x\text{-axis, not necessarily at the Origin.}\)

\(\displaystyle \text{Both [2] and [3] have one }x\text{-intercept.}\)

 

[JeffM]

Soroban has given you the right answer, but it may help to supplement a little, which I was trying to do socratically. But let's dispense with that.

In a polynomial of degree n, the term ax[sup:31a2fziv]n[/sup:31a2fziv] dominates for all x greater than one critical value and for all x less than a critical value. (The two critical values may or may not be the same.) That is, you can get an idea of what the graph looks like for positive x with a very large absolute value and for negative x with a very large absolute value simply by thinking about ax[sup:31a2fziv]n[/sup:31a2fziv] and forgetting about the rest of the equation. So, in the quadratic, the arms of the parabola reach upward if a > 0 and reach downward if a < 0. This trick of course does not help you understand what is happening within the critical values.

Now the quadratic formula tells you about the roots of any quadratic, that is where the quadratic has a value of zero and the function intercepts the x axis. The portion of the formula that is b[sup:31a2fziv]2[/sup:31a2fziv] - 4ac is called the discriminant. If it is positive, there are TWO real roots and so TWO intercepts. If it is zero, there is ONE real root. If it is negative, there are ZERO real roots.

So you can use those two facts to eliminate at least some of the parabolas. If there is more than one that has upward arms and just one x intercept, you have to have additional information that I do not think you have given us.

 

[mmm4444bot]

the correct equation in the problem is axsquared + bx +c, where bsquared + 4ac= 0

Twice, we've asked you to type polynomials using math notation and the caret symbol instead of English words.

The caret symbol ^ is a shifted six, on most keyboards. Please use it like this:

y = ax^2 + bx + c, where b^2 - 4ac = 0