Describe the y-intercept and end behavior of the following graph:

If you were to multiply the three factors using the distributive property, you would find that this function is a third degree polynomial because the term with the dependent variable raised to highest power would be the term #x^3#.

Since the term with the dependent variable raised to the highest power has a positive coefficient #(+1)# and an odd power #(3)#, end behavior of #f(x)# when #x# becomes more negative will move in the direction of #-infty# and when #x# becomes more positive will move in the direction of #+\infty#.
end behavior: #(-\infty, -\infty)#, #(\infty, \infty)#

To find the zeros, or x-intercepts, of the function, set #f(x) = 0# and solve for #x#. Using the zero product property, we know that #f(x) = 0# when any one of the three factors, #(x-4)#, #(x-1)#, or #(x+3)# is equal to #0#. Therefore:
x-intercepts: #(4,0)# , #(1,0)#, and #(-3,0)#.

To find the y-intercept of the function, set #x = 0# and solve for #f(x)#:
#f(x) = (x - 4)(x - 1)(x + 3)#
#f(0) = ((0) - 4)((0) - 1)((0) + 3)#
#f(0) = (- 4)(- 1)(3)#
#f(0) = 12#
y-intercept: #(0,12)#

The end behavior of a polynomial function is the behavior of the graph of f(x) as x approaches positive infinity or negative infinity.

The degree and the leading coefficient of a polynomial function determine the end behavior of the graph.

The leading coefficient is significant compared to the other coefficients in the function for the very large or very small numbers. So, the sign of the leading coefficient is sufficient to predict the end behavior of the function.

Degree	Leading Coefficient	End behavior of the function	Graph of the function
Even	Positive	f(x)→+∞, as x→−∞f(x)→+∞, as x→+∞	Example: f(x)=x2
Even	Negative	f(x)→−∞, as x→−∞f(x)→−∞, as x→+∞	Example: f(x)=−x2
Odd	Positive	f(x)→−∞, as x→−∞f(x)→+∞, as x→+∞	Example: f(x)=x3
Odd	Negative	f(x)→+∞, as x→−∞f(x)→−∞, as x→+∞	Example: f(x)=−x3

To predict the end-behavior of a polynomial function, first check whether the function is odd-degree or even-degree function and whether the leading coefficient is positive or negative.

Example:

Find the end behavior of the function x4−4x3+3x+25.

The degree of the function is even and the leading coefficient is positive. So, the end behavior is:

f(x)→+∞, as x→−∞f(x)→+∞, as x→+∞

The graph looks as follows:

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How to Determine End Behavior & Intercepts to Graph a Polynomial Function

Step 1: Identify the x-intercept(s) of the function by setting the function equal to 0 and solving for x. If they exist, plot these points on the coordinate plane.

Step 2: Identify the y-intercept of the function by plugging 0 into the function. Plot this point on the coordinate plane.

Step 3: Identify the end behavior of the function by looking at the leading term. The parity of the exponent and the sign of the coefficient determines the end behavior of the function as a whole. Put arrows in the respective quadrants that the function extends off to.

Step 4: Connect the arrows and points together to create a rough sketch of what the polynomial function looks like.

How to Determine End Behavior & Intercepts to Graph a Polynomial Function: Vocabulary

End behavior: The end behavior of a polynomial function describes how the graph behaves as {eq}x {/eq} approaches {eq}\pm\infty. {/eq} We can determine the end behavior by looking at the leading term (the term with the highest {eq}n {/eq}-value for {eq}ax^n {/eq}, where {eq}n {/eq} is a positive integer and {eq}a {/eq} is any nonzero number) of the function. Depending on the sign of the coefficient ({eq}a {/eq}) and the parity of the exponent ({eq}n {/eq}), the end behavior differs;

Note that "{eq}\longrightarrow {/eq}" is a way to abbreviate "approaches." Since this is how we can describe the end behavior for all polynomial functions, then if we know how one specific function behaves with the respective signs/parities, we know how all polynomial functions look. In this case, think of the end behavior of {eq}x^2, \ -x^2, \ x^3, \ -x^3, {/eq} respectively:

Intercepts: The y-intercept of a polynomial function is the point {eq}(0,b) {/eq} that crosses through the y-axis. The x-intercept(s) of a polynomial function (if they exist) are the point(s) {eq}(a,0) {/eq} that cross through the x-axis. If we have an odd number of copies of a certain x-intercept, then the graph will pass through the intercept (think of how {eq}x^3 {/eq} behaves at its intercept). If we have an even number of copies of a certain x-intercept, then the graph will just touch the intercept (think of how {eq}x^2 {/eq} behaves at its intercept).

By knowing the end behavior and intercepts of the function only, we can determine roughly what the function looks like on the coordinate plane.

We will now go through two examples step-by-step.

How to Determine End Behavior & Intercepts to Graph a Polynomial Function: Example 1

Graph the following function by determining the end behaviors and intercepts from the equation: {eq}f(x)=(x-2)^2(x+1)(x+5). {/eq}

Step 1: Identify the x-intercept(s).

To identify the x-intercept(s), we need to set the equation equal to 0, i.e., let {eq}f(x)=0=(x-2)^2(x+1)(x+5). {/eq} We then have three different possibilities:

1. {eq}\begin{align} (x-2)^2&=0\\ x-2&=0\\ x&=2 \end{align} {/eq}

2. {eq}\begin{align} x+1&=0\\ x&=-1 \end{align} {/eq}

3. {eq}\begin{align} x+5&=0\\ x&=-5 \end{align} {/eq}

This means the x-intercepts are {eq}\mathit{(2,0), \ (-1,0), \ (-5,0)} {/eq}:

Step 2: Identify the y-intercept.

To identify the y-intercept, we need to plug in 0 into the function, i.e., {eq}f(0)=(0-2)^2(0+1)(0+5)=(-2)^2(1)(5)=(4)(1)(5)=20. {/eq} So, the y-intercept is {eq}\mathit{(0,20)} {/eq}:

Step 3: Identify the end behavior.

We need to expand the function to determine what the leading term is:

{eq}\begin{align} f(x)&=(x^2-4x+4)(x+1)(x+5)\\ &=(x^3-3x^2+4)(x+5)\\ &=x^4+2x^3-15x^2+4x+20 \end{align} {/eq}

Looking for the term with the highest exponent, the leading term is {eq}x^4 {/eq}. Since this has a positive coefficient and an even exponent, then as {eq}x {/eq} approaches both {eq}\pm\infty, \ f(x) {/eq} approaches {eq}\infty, {/eq} meaning, roughly, the graph extends up at both ends of the graph:

Step 4: Connect the graph.

We connect the graph by observing the following:

• We start at the top left corner of the coordinate plane and cross the x-axis through the point {eq}(-5,0) {/eq} since we only have one copy of this intercept.

• We need to turn around and cross the x-axis again through the point {eq}(-1,0) {/eq} since we only have one copy of this intercept.

• We need to cross the y-axis through the point {eq}(0,20) {/eq}.

• We are just going to touch the point {eq}(2,0) {/eq} since we have an even number of copies of this intercept, and we will extend into the top right corner of the coordinate plane.

This produces the following rough sketch:

Note that this isn't too far off from the actual shape of the graph of {eq}f(x)=(x-2)^2(x+1)(x+5): {/eq}

How to Determine End Behavior & Intercepts to Graph a Polynomial Function: Example 2

Graph the following function by determining the end behaviors and intercepts from the equation: {eq}f(x)=-(x-3)^2(x-1)^2(x+2). {/eq}

Step 1: Identify the x-intercept(s).

To identify the x-intercept(s), we need to set the equation equal to 0, i.e., let {eq}f(x)=0=-(x-3)^2(x-1)^2(x+2). {/eq} We then have three different possibilities:

1. {eq}\begin{align} (x-3)^2&=0\\ x-3&=0\\ x&=3 \end{align} {/eq}

2. {eq}\begin{align} (x-1)^2&=0\\ x-1&=0\\ x&=1 \end{align} {/eq}

3. {eq}\begin{align} x+2&=0\\ x&=-2 \end{align} {/eq}

This means the x-intercepts are {eq}\mathit{(3,0), \ (1,0), \ (-2,0)} {/eq}:

Step 2: Identify the y-intercept.

To identify the y-intercept, we need to plug in 0 into the function, i.e., {eq}f(0)=-(0-3)^2(0-1)^2(0+2)=-(-3)^2(-1)^2(2)=-(9)(1)(2)=-18. {/eq} So, the y-intercept is {eq}\mathit{(0,-18)} {/eq}:

Step 3: Identify the end behavior.

We need to expand the function to determine what the leading term is:

{eq}\begin{align} f(x)&=-(x^2-6x+9)(x^2-2x+1)(x+2)\\ &=-(x^4-8x^3+22x^2-24x+9)(x+2)\\ &=-(x^5-6x^4+6x^3+20x^2-39x+18)\\ &=-x^5+6x^4-6x^3-20x^2+39x-18 \end{align} {/eq}

Looking for the term with the highest exponent, the leading term is {eq}-x^5 {/eq}. Since this has a negative coefficient and an odd exponent, then as {eq}x {/eq} approaches {eq}-\infty, \ f(x) {/eq} approaches {eq}\infty, {/eq} and as {eq}x {/eq} approaches {eq}\infty, \ f(x) {/eq} approaches {eq}-\infty. {/eq} This means, roughly, the graph extends off up to the left and down to the right:

Step 4: Connect the graph.

We connect the graph by observing the following:

• We start at the top left corner of the coordinate plane and cross the x-axis through the point {eq}(-2,0) {/eq} since we only have one copy of this intercept.

• We need to cross the y-axis through the point {eq}(0,-18) {/eq}.

• We touch the point {eq}(1,0) {/eq} on the x-axis, since we have two copies of this intercept.

• We need to turn around and "touch" the point {eq}(3,0) {/eq} on the x-axis, since we also have two copies of this intercept, and we will extend into the bottom right corner of the coordinate plane.

This produces the following rough sketch:

Note that this isn't too far off from the actual shape of the graph of {eq}f(x)=-(x-3)^2(x-1)^2(x+2): {/eq}

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