Advanced Functions Question

[Otis]

… the x-value … is the changing point for the positive and negative intervals

Hello. If by "intervals" you mean "rates of change", then I agree. The x-value that lies halfway between the two given roots is the point where the rate of change goes from positive to negative OR goes from negative to positive.

Graphically speaking, the value halfway between the roots is the x-coordinate of the turning point (which we call the 'vertex' of the parabola). But I'm going to ignore graphs altogether because your question is about how to use the roots to help find where the rate of change is positive or negative. (If we have a graph, we don't need to use the roots; we just look at the graph.)

I'm assuming that we've only been given the two roots and the sign of the leading coefficient.

For example, using a>0 and two given roots -14 and 5, we find the value halfway between the roots (their average):

x = (root1 + root2)/2

x = (-14 + 5)/2 = -9/2

So, the x-coordinate at the turning point is -9/2. Now we can evaluate the function at the turning point and to the right of the turning point -- to see whether y increases or decreases. I'll use -4.5 and -4.0 as the two test values. Because we're not asked to evaluate the function at any point, we can use a=+1.

y = (+1)(x - root1)(x - root2)

Here is the first test output:

y = (-4.5 + 14)(-4.5 - 5) = -90.25

Here is the second test output, after increasing x by 0.5:

y = (-4.0 + 14)(-4.0 - 5) = -90.00

We see that y increased (by 0.25 units), so the rate of change is positive to the right of the vertex and negative to the left. (If y had decreased, instead, then the rate of change is negative to the right of the vertex and positive to the left.)

?

 

[Mumtaaz]

Hello. If by "intervals" you mean "rates of change", then I agree. The x-value that lies halfway between the two given roots is the point where the rate of change goes from positive to negative OR goes from negative to positive.

Graphically speaking, the value halfway between the roots is the x-coordinate of the turning point (which we call the 'vertex' of the parabola). But I'm going to ignore graphs altogether because your question is about how to use the roots to help find where the rate of change is positive or negative. (If we have a graph, we don't need to use the roots; we just look at the graph.)

I'm assuming that we've only been given the two roots and the sign of the leading coefficient.

For example, using a>0 and two given roots -14 and 5, we find the value halfway between the roots (their average):

x = (root1 + root2)/2

x = (-14 + 5)/2 = -9/2

So, the x-coordinate at the turning point is -9/2. Now we can evaluate the function at the turning point and to the right of the turning point -- to see whether y increases or decreases. I'll use -4.5 and -4.0 as the two test values. Because we're not asked to evaluate the function at any point, we can use a=+1.

y = (+1)(x - root1)(x - root2)

Here is the first test output:

y = (-4.5 + 14)(-4.5 - 5) = -90.25

Here is the second test output, after increasing x by 0.5:

y = (-4.0 + 14)(-4.0 - 5) = -90.00

We see that y increased (by 0.25 units), so the rate of change is positive to the right of the vertex and negative to the left. (If y had decreased, instead, then the rate of change is negative to the right of the vertex and positive to the left.)

?

This method is very helpful, I just have a couple of questions, Did you use -4.5 and -4.0 because they are close to -5? Also, is it possible to use instantaneous rate of change to explain the same process?

 

[Otis]

… Did you use -4.5 and -4.0 because they are close to -5? …

Nope. The number -5 is not part of my example. My example uses +5 and -14 as two given roots.

I used -4.5 for x because that's the x-value halfway between the two given roots.

I used -4.0 because I needed to increase x a little bit, to get a test value to the right of the vertex. I chose to increase -4.5 by 0.5, but the increase amount is not important; I could have increased by 0.1, instead (using -4.5 and -4.4).

… is it possible to use instantaneous rate of change to explain the same process?

No. My process to answer the exercise in this thread assumes that the student has not yet learned how to calculate the instantaneous rate of change. That's a calculus method.

?

 

[Otis]

Oops, I need to edit post #21. I went too fast and completely forgot to state that a>0 in my example. As Jomo said earlier, we need to know also the sign of the leading coefficient (a) because IF that sign is not given, then having only the two roots is not enough information for a single answer. In such a case, we would need to provide a conditional answer listing both possibilities (a>0 and a<0).

Also, I'm using test values of y -- not the actual values of y (because we don't know the actual function). For my method, I use only a=1 or a=-1, in the test-value formula. That shortcut doesn't matter because the sign of the rate of change doesn't depend on the value of a, only its sign.

y = (a)(x - root1)(x - root2)

So, in my example, I need to edit the post to show:

y = (+1)(x - root1)(x - root2)

The actual function -- which is totally unknown to us -- could have any Real number (other than zero) for the leading coefficient a. My method doesn't care about the value of a, only its sign.

If we didn't know the sign of a, in my example, then we could state the answer like this:

Case a > 0
The rate of change is negative on (-∞, -9/2) and positive on (-9/2, ∞)

Case a < 0
The rate of change is positive on (-∞, -9/2) and negative on (-9/2, ∞)

I hope my sloppiness hasn't confused you. (I may have taken too much medicine, tonight.) Perhaps, I will post a couple more examples, tomorrow. The line through the test points creates a secant line. It's the slope of that line that corresponds to an average rate of change, so we only need to consider the sign of the slope of our test secant line. (We don't need to calculate a number). We use what we learned from working with lines: when x increases, a decreasing y means negative slope and an increasing y means positive slope.

?

 

[Mumtaaz]

Nope. The number -5 is not part of my example. My example uses +5 and -14 as two given roots.

I used -4.5 for x because that's the x-value halfway between the two given roots.

I used -4.0 because I needed to increase x a little bit, to get a test value to the right of the vertex. I chose to increase -4.5 by 0.5, but the increase amount is not important; I could have increased by 0.1, instead (using -4.5 and -4.4).


No. My process to answer the exercise in this thread assumes that the student has not yet learned how to calculate the instantaneous rate of change. That's a calculus method.

?

Oh okay , thanks so much for the help!

 

[Otis]

, thanks

You're welcome. Will you be taking a calculus class, in the future? Calculus uses a limit of changing secant line slopes, when defining the instantaneous rate of change. (I chose a secant-line approach because you're in precalculus.)

If you won't be taking calculus, then I can show another way to answer the exercise as given -- a method that doesn't require knowing anything about the leading coefficient a. It requires thinking about the two basic shapes of a parabola (more of an intermediate-algebra approach.)

I'm still not sure what your instructor had in mind. Maybe any reasoning is acceptable.

?

 

[Mumtaaz]

You're welcome. Will you be taking a calculus class, in the future? Calculus uses a limit of changing secant line slopes, when defining the instantaneous rate of change. (I chose a secant-line approach because you're in precalculus.)

If you won't be taking calculus, then I can show another way to answer the exercise as given -- a method that doesn't require knowing anything about the leading coefficient a. It requires thinking about the two basic shapes of a parabola (more of an intermediate-algebra approach.)

I'm still not sure what your instructor had in mind. Maybe any reasoning is acceptable.

?

No, I don't plan to take one anytime soon, I've been thinking I should in uni. If you can show me another way that would be great too!

 

[Mumtaaz]

Oops, I need to edit post #21. I went too fast and completely forgot to state that a>0 in my example. As Jomo said earlier, we need to know also the sign of the leading coefficient (a) because IF that sign is not given, then having only the two roots is not enough information for a single answer. In such a case, we would need to provide a conditional answer listing both possibilities (a>0 and a<0).

Also, I'm using test values of y -- not the actual values of y (because we don't know the actual function). For my method, I use only a=1 or a=-1, in the test-value formula. That shortcut doesn't matter because the sign of the rate of change doesn't depend on the value of a, only its sign.

y = (a)(x - root1)(x - root2)

So, in my example, I need to edit the post to show:

y = (+1)(x - root1)(x - root2)

The actual function -- which is totally unknown to us -- could have any Real number (other than zero) for the leading coefficient a. My method doesn't care about the value of a, only its sign.

If we didn't know the sign of a, in my example, then we could state the answer like this:

Case a > 0
The rate of change is negative on (-∞, -9/2) and positive on (-9/2, ∞)

Case a < 0
The rate of change is positive on (-∞, -9/2) and negative on (-9/2, ∞)

I hope my sloppiness hasn't confused you. (I may have taken too much medicine, tonight.) Perhaps, I will post a couple more examples, tomorrow. The line through the test points creates a secant line. It's the slope of that line that corresponds to an average rate of change, so we only need to consider the sign of the slope of our test secant line. (We don't need to calculate a number). We use what we learned from working with lines: when x increases, a decreasing y means negative slope and an increasing y means positive slope.

?

Ah don't worry I'm not too confused just had to read it more than twice because my brain lags, so just to clarify, the slope is to better illustrate how the y is increasing and so it is a positive slope and a positive rate of change in this example, right?

 

[Steven G]

To show us that you completely understand this please go back to your two graphs and

1) one by one highlight and label where where the function has a positive rate of change and a negative rate of change.

2) Then note that in each graph the point where the rate of change changed its sign happens to be at the vertex.

3) Then notice that the x-value of the vertex is the average of the two roots.

I think that the ordering is important as you get to see the whole picture from beginning to the end.