two distinct zeros

[sajoshi21]

Since you guys are so much more helpful than our book maybe you can guide me with the other problem as well.

Sketch a cubic function y=p(x) with two distinct zeros at x=2 and x=5 and has a local maximum located at x=5. Hint: you will have one double zero

No where in our material does it show how to begin to solve this. Thanks in advance

 

[mmm4444bot]

Math texts do not provide step-by-step instructions for every exercise. You are expected to apply what you have learned from previous lessons and examples to new situations.

You ought to have learned that the graph of a function touches the x-axis at its Real zeros. We call these points x-intercepts.

When a function has a double zero, then the graph does not cross over the x-axis at that point, but rather it turns around.

Look at the graphs of cubic polynomials below.


This graph has three x-intercepts, so we know that the cubic function has three distinct Real zeros.

We see the zeros at -2, 0, and 1.






This graph shows only two x-intercepts. We see that the curve does not cross the x-axis at (2,0) -- it turns around there.

Hence, we know that this cubic function has a double zero at x=2.




Is this enough information for you to try your exercise?

 

[DrPhil]

Since you guys are so much more helpful than our book maybe you can guide me with the other problem as well.

Sketch a cubic function y=p(x) with two distinct zeros at x=2 and x=5 and has a local maximum located at x=5. Hint: you will have one double zero

No where in our material does it show how to begin to solve this. Thanks in advance

If you call the unknown root "r", then the factored form of the cubic equation is
\(\displaystyle \displaystyle y = (x - 2)(x - 5)(x - r)\)

My first thought to deal with the local maximum at x=5 was to use calculus, BUT I see that this is posted in the algebra folder so you probably wouldn't know what I was talking about. So different approach. If you sketch the function at x=5, if it has a local maximum there it will have a vertex similar to an inverted parabola. Since you also know the value of the function is 0 when x=5, you will see that the function does not pass through y=0, but just kisses the x-axis. That is only possible if there are two roots at that point, and r=5. Does that help to make your sketch?