Graphing Polynomial Functions - Algebra 2

The graph of a quadratic function has an -intercept at any point at which , so, first, set the quadratic expression equal to 0:

The number of -intercepts of the graph is equal to the number of real zeroes of the above equation, which can be determined by evaluating the discriminant of the equation, . Set , and evaluate:

The discriminant is equal to zero, so the quadratic equation has one real zero, and the graph of has exactly one -intercept.

The parabola of an equation of the form is vertical, and faces upward or downward depending entirely on the sign of , the coefficient of . This coefficient, , is negative; the parabola is concave downward.

The graph of an equation of the form

is a horizontal parabola. Whether it is concave to the left or to the right depends on the sign of . Since , a negative number, the parabola is concave to the left.

The -intercept of the graph of a function is the point at which it crosses the -axis; its -coordinate is 0, so its -coordinate is .

,

so, by setting ,

,

making the -intercept.

We have the following answer choices.

The first equation is a cubic function, which produces a function similar to the graph. The second equation is quadratic and thus, a parabola. The graph does not look like a prabola, so the 2nd equation will be incorrect. The third equation describes a line, but the graph is not linear; the third equation is incorrect. The fourth equation is incorrect because it is an exponential, and the graph is not an exponential. So that leaves the first equation as the best possible choice.

Starting with

moves the parabola by units to the right.

Similarly moves the parabola by units to the left.

Hence the correct answer is option .

The highest exponent of the variable term is two (). This tells that this function is quadratic, meaning that it is a parabola.

The graph below will be the answer, as it shows a parabolic curve.

To find where the graph crosses the horizontal axis, we need to set the function equal to 0, since the value at any point along the axis is always zero.

To find the possible rational zeroes of a polynomial, use the rational zeroes theorem:

Our constant is 10, and our leading coefficient is 1. So here are our possible roots:

Let's try all of them and see if they work! We're going to substitute each value in for using synthetic substitution. We'll try -1 first.

Looks like that worked! We got 0 as our final answer after synthetic substitution. What's left in the bottom row helps us factor down a little farther:

We keep doing this process until is completely factored:

Thus, crosses the axis at .

crosses the axis when equals 0. So, substitute in 0 for :

The zeros of the parabola are at and , so when placed into the formula

,

each of their signs is reversed to end up with the correct sign in the answer. The coefficient can be found by plugging in any easily-identifiable, non-zero point to the above formula. For example, we can plug in which gives

When determining the maximum number of turns a polynomial function might have, one must remember:

Max Number of Turns for Polynomial Function = degree - 1

First, we must find the degree, in order to determine the degree we must put the polynomial in standard form, which means organize the exponents in decreasing order:

Now that f(x) is in standard form, the degree is the largest exponent, which is 8.

We now plug this into the above:

Max Number of Turns for Polynomial Function = degree - 1

Max Number of Turns for Polynomial Function = 8 - 1

which is 7.

The correct answer is 7.

In order to determine the end behavior of a polynomial function, it must first be rewritten in standard form. Standard form means that the function begins with the variable with the largest exponent and then ends with the constant or variable with the smallest exponent.

For f(x) in this case, it would be rewritten in this way:

When this is done, we can see that the function is an Even (degree, 4) Negative (leading coefficent, -3) which means that both sides of the graph go down infinitely.

In order to answer questions of this nature, one must remember the four ways that all polynomial graphs can look:

Even Positive:

Even Negative:

Odd Positive:

Odd Negative:

The way to figure out this problem is by understanding behavior of polynomials.

The sign that occurs before the is positive and therefore it is understood that the function will open upwards. the "8" on the function is an even number which means that the function is going to be u-shaped. The only answer choice that fits both these criteria is:

When we look at the function we see that the highest power of the function is a 3 which means it is an "odd degree" function. This means that the right and left side of the function will approach opposite directions. *Remember O for Odd and O for opposite.

In this case we also have a negative sign associated with the highest power portion of the function - this means that the function is flipped.

Both of these combine to make this an "odd negative" function.

Odd negative functions always have the right side of the function approaching down and the left side approaching up.

We represent this mathematically by saying that as x approaches negative infinity (left side), the function will approach positive infinity:

...and as x approaches positive infinity (right side) the function will approach negative infinity:

Then set each factor equal to zero, if any of the ( ) equal zero, then the whole thing will equal zero because of the zero product rule.

As a polynomial function, the graph of is continuous. By the Intermediate Value Theorem, if or , then there must exist a value such that .

Set and . It is not true that , so the Intermediate Value Theorem does not prove that there exists such that . However, it does not disprove that such a value exists either. For example, observe the graphs below:

Both are polynomial graphs fitting the given conditions, but the only the equation graphed at right has a zero on .

As a polynomial function, the graph of is continuous. By the Intermediate Value Theorem (IVT), if or , then there must exist a value such that .

Setting , and examining the first condition, the above becomes:

if , then there must exist a value such that - or, restated, must have a zero on the interval . Since , . the condition holds, and by the IVT, it follows that has a zero on .

As a polynomial function, the graph of is continuous. By the Intermediate Value Theorem, if or , then there must exist a value such that .

Setting , and , this becomes: If or , then there must exist a value such that - that is, must have a zero on .

However, the question is asking us to use the converse of this statement, which is not true in general. If has a zero on , it does not necessarily follow that or - specifically, with , it does not necessarily follow that . A counterexample is the function shown below, which fits the conditions of the problem but does not have a negative value for :

The answer is false.

As a polynomial function, the graph of is continuous. By the Intermediate Value Theorem, if or , then there must exist a value such that .

Setting and , assuming for now that , and looking only at the second condition, this statement becomes: If , then there must exist a value such that - or, equivalently, must have a zero on .

However, the conclusion of this statement is false: has no zeroes at all. Therefore, is false, and has no negative values for any . By similar reasoning, has no negative values for any . Therefore, by the IVT, by way of its contrapositive, we have proved that is positive everywhere.