## End Behavior, Degree, and Leading Coefficient

This post about end behavior, degree, and leading coefficient of a polynomial function is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

Consider the function

$\fn_jvn {\color{DarkBlue} g(x)=a(x-3)(x+4)^c}$

where a and c are integers and are constants and c is positive. The the graph y = g(x) goes toward negative infinity at both negative and positive extremes of x. Is positive or negative? Is c even or odd?

## Solution

Let’s deal with c first.

### Sidebar: even and odd polynomial functions

The end behavior of a polynomial function with an even degree is in the same direction — either toward negative infinity or toward positive infinity — for both extremes of x. Think, for example, of  $\inline \fn_jvn {\color{DarkBlue} y=x^2}$. It goes up toward positive infinity as the absolute value of x increases, whether positive or negative. So do $\inline \fn_jvn y=x^4$, $\inline \fn_phv y=x^6$, etc.

 $\inline \fn_jvn {\color{DarkBlue} y=x^2}$$\inline \fn_jvn {\color{DarkBlue} y=x^2}$ $\inline \fn_jvn y=x^4$$\inline \fn_jvn y=x^4$ $\inline \fn_jvn y=x^6$$\inline \fn_jvn y=x^6$

If you multiply any of those expressions by a leading coefficient of -1, or any negative number, then end behavior goes to negative infinity for both extremely negative and extremely positive values of x.

On the other hand, the end behavior of a polynomial with an odd degree is in opposite directions for extremely negative and extremely positive values of x. Think of $\inline \fn_jvn {\color{DarkBlue} y=x}$, $\inline \fn_jvn y=x^3$, etc.

 $\inline \fn_jvn {\color{DarkBlue} y=x}$$\inline \fn_jvn {\color{DarkBlue} y=x}$ $\inline \fn_jvn y=x^3$$\inline \fn_jvn y=x^3$ $\inline \fn_jvn y=x^5$$\inline \fn_jvn y=x^5$

So far, I’ve been talking about a monomial — $\inline \fn_jvn {\color{DarkBlue} y=x}$ or $\inline \fn_jvn {\color{DarkBlue} y=x^2}$ or y equals x to some other power. What happens in the more general case, when you get a polynomial with several terms, for example $\inline \fn_jvn y=x^5+4x^4-2x^2+1$?

 $\inline \fn_jvn y=x^5+4x^4-2x^2+1$$\inline \fn_jvn y=x^5+4x^4-2x^2+1$

When a polynomial contains terms of lower degree than the leading term, you get some noise, but end behavior is still determined by the highest-degree term. Note that as x gets very negative or very positive, the graph of $\inline \fn_jvn y=x^5+4x^4-2x^2+1$ starts to look more like the graph of $\inline \fn_jvn y=x^5$.

### Back to the Question

The function in question, $\inline \fn_jvn g(x)=a(x-3)(x+4)^c$, has degree c + 1: The c comes from the power c on the term x + 4, and the +1 comes from the implied power of 1 on the term x – 3, for a total power of x  equal to c + 1. Thus c + 1 is the degree of function g. And you’re told that this function’s end behavior is to approach negative infinity at both extremes of x. As shown above, a polynomial that goes in the same direction — either up or down — for both extremely positive and extremely negatives values of x has an even degree. That means c+1 is even, so c is odd.

What about a? As shown above, a polynomial with even degree, like this one, that goes down at both extremes of x starts with a negative number. So a is negative.

This question is similar to question number 17 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

# How Many x-Intercepts?

## Question

What is the maximum number of zeros (also called x-intercepts) the function $f(x)=x^4-2x+c$   may have? What is the minimum? c, a constant, may be any real number.

## Solution

The maximum number zeros of a polynomial function is equal to the function’s degree. The function $f(x)=x^4-2x+c$ is fourth degree, so it may have up to four zeros. A fourth-degree function may look like this:

Four is the max. Will the function $f(x)=x^4-2x+c$ necessarily have four? Let’s try to narrow down the possibilities.

Try a graph. You can graph this function if you assign a value to c. Let’s try $c=0$, so that the function is $y=x^4-2x$. That graphs up like this:

That gives you two x-intercepts. but with a different value of c could this function may have more x-intercepts — or fewer?

At the x-intercepts, the y-value is 0:

$x^4-2x+c=0$

That means

$x^4=2x-c$

That means that for this function, at the zeros, the fourth-degree monomial $x^4$  equals the linear function $2x-c$. Let’s graph these two pieces separately for $c=0$ , the value of c that we’ve seen gives two x-intercepts.

These two pieces intersect in to placesone for each of the two zeros. Are any other points of intersection are possible? On the left-hand side, the fourth-degree term just keeps going up while the line just keeps going down, so there are no more points of intersection on the left. On the right-hand side to the right of both points of intersection, the fourth-degree term is always steeper than the line and is therefore always greater, so there are no more points of intersection on the right. Remember that each point of intersection on this graph corresponds to a zero on the function $f(x)=x^4-2x+c$, so that means no other zero is possible for the function.

But that doesn’t cover all possibilities, because we considered only one value for c. What if c takes a different value? Then the line slides up or down. If $c>0$ , the line slides up and there will two points of intersection, one at which x is less than zero and one in which x is greater than zero.

Slide it down just a little from $c=0$ and there are still two points of intersection, this time both positive. Slide it farther and eventually it will not intersect the fourth-degree curve at all.

For even higher values of c – meaning the line slides farther down on the y-axis – the function $x^2-2x+c$ has no zeros.

So for the function $f(x)=x^4-2x+c$, the minimum number of x-intercepts is zero and the maximum number is two.

# Maximum Value on an Interval

## Question

If a toy car moves according to the equation $v(t)=2/3t^3-4t^2+6t+2$, where t represents time and v represents velocity, what is the toy car’s maximum acceleration on the interval [0, 5]?

## Solution

Acceleration is the first derivative of velocity. So if $v(t)=2/3t^3-4t^2+6t+2$ represents the toy car’s velocity, then $a(t)=2t^2-8t+6$ represents its acceleration.

### How to find the maximum?

The maximum on the interval could be a local maximum somewhere within the interval, or it could be a value at one end or the other. You can calculate all three and compare them. To get started a sketch may help you get a sense of what the curve looks like.

To make a quick sketch of $a(t)=2t^2-8t+6$, factor the trinomial to $2(t-3)(t-1)$. That tells you the x-intercepts: $(3,&space;0)$ and $(5,&space;0)$. And the 6 at the end of $2t^2-8t+6$ tells you that the y-intercept is $(0,&space;6)$. Those three points are enough for a crude sketch. I’ve extended the domain to 5 on the right, because that’s the interval that’s given. I think the y-value is higher at $x=5$ than it is at $x=0$, because 5 is farther from 2, which appears to be the vertex.

As you can see from either the sketch or the equation (the leading coefficient is positive), this parabola opens upward. Setting the acceleration function’s first derivative equal to zero will tell you where the vertex is – but as you can see that vertex is a minimum, not a maximum, so that is not the route to pursue to find the maximum.

### Where the first derivative is zero or at  the interval’s end

Since the vertex is the only turning point on the interval, the maximum value on the interval must be at one of the interval’s ends. If the sketch is accurate, the acceleration function, $a(t)=2t^2-8t+6$, should be at its maximum in the interval at $t=5$.

To confirm, you may want to calculate acceleration at several key points. To find acceleration at the left-hand end of the interval, $t=0$, is easy: Substitute 0 for t and you get $a(0)=6$.

Next, try the local minimum. To find the y-value there, you first have to find the t-value. To do that, differentiate $a(t)=2t^2-8t+6$ and get $a’(t)=4t-8$. Set that to 0 to get

$4t-8=0$

$t=2$

(Alternatively, since the acceleration function is second degree, if you represent its equation as $at^2+bt+c$, you can find its vertex by setting t equal to $-b/(2a)$. For the equation $a(t)=2t^2-8t+6$$a=2$ , $b=-8$, and $c=6$, so $-b/(2a)=2$.)

$a(2)=2(2^2)-8(2)+6=-2$

Finally, to find $a(5)$ you plug 5 into $a(t)=2t^2-8t+6$, which is a bit more work than plugging in 0, but still not a big deal. Then

$a(5)=2(5^2)-8(5)+6=16$

The results:

 t A(t) 0 6 2 -2 5 16

As the crude sketch indicated, the highest value is at $t=5$. There $a(t)=16$, and that is the answer. If you can work with a calculator and are not limited to what you can sketch by hand, you may get a better picture, like this: