create quadratic eqn w/ min pt, 2 pos real zeroes, y-int =2

[falloutatthedisco]

hi,

how would I create a quadratic function with a minimum point, two positive real zeros, and a y-intercept of 2?

thank you!

 

[mmm4444bot]

Re: create an equation...

Hi Fallout:

I cannot answer your question, yet, because I do not know how you would create a quadratic function for anything. In other words, I don't know what kind of help you need to answer this exercise.

I would like to know whether or not you have learned about the "discriminant", or about translations (such as those involving quadratic functions) to shift graphs around.

For examples, have you learned about the relationship between the sign of b^2-4ac and the number of real solutions for ax^2+bx+c=0; do you know what happens to the graph of ax^2+bx+c if somebody alters the value of c; have you learned about shifting graphs and seeing how the algebraic expressions change from one graph to another; if you are not into these translations, yet, then are you just studying parabolas; are you studying the steps to change three (x,y) coordinates from a parabolic graph into the algebraic definition for that quadratic function.

Does any of this stuff ring a bell?

If it does not, then please tell me more about why you want to discover an algebraic definition for such a particular quadratic function.

Here's an example of what the answer looks like:

f(x) = ax^2 + bx + c

where a, b, and c are replaced with three real numbers.

Cheers,

~ Mark

 

[soroban]

Re: create an equation...

Hello, falloutatthedisco!

We can talk our way through it . . . step by step.

Create a quadratic function with a minimum point, two positive real zeros, and a y-intercept of 2

Consider what this function demands . . .

"Quadratic function": the equation is of degree two; the graph is a parabola.

"with a minimum point": the minimum occurs at the vertex; it is an up-opening parabola: \(\displaystyle \cup\)

"two positive real zeros": the two x-intercepts are to the right of the origin.

"and a y-intercept of 2": the graph crosses the y-axis at (0,2)


The graph looks like this:

       |
     * |                   *
       |
      *|                  *
      2o                 *
       | *             *
     --+----o-------o--------
       |        *

As Mark pointed out, we don't know what methods you have learned
. . for completing the solution.

 

[falloutatthedisco]

Re: create an equation...

I am supposed to find the equation using real zeroes of the function, or I can use discriminants, because yes, I have learned how to use those.

Thank you for your help.

 

[stapel]

how would I create a quadratic function with a minimum point, two positive real zeros, and a y-intercept of 2?

For a parabola to have a "minimum" point, what sign must the quadratic have on the squared term?

For a quadratic to have a zero of x = a, then it must have a factor of x - a. (Think about solving by factoring, if you're not sure about this.) If you need two positive roots, then what can you say about the factors?

For the y-intercept of y = ax[sup:1g8pzhm3]2[/sup:1g8pzhm3] + bx + c to be zero, what must be the value of "c"?

For factors (x - r)(x - s), what would be a useful value for the product of r and s?

If the above is not sufficient to help you to a solution, kindly please reply with a clear listing of your answers to each question, along with any other thoughts or efforts. Thank you!

Eliz.