High School Algebra II Unlocked (2016)

Lesson 1.4. Graphing Polynomials (Beyond Quadratics)

The vertex form of a quadratic function, f(x) = a(x − h)² + k, tells you that its graph is the result of shifting the graph of f(x) = ax² a total of h units to the right and k units up. If h is negative, the graph is shifted to the left, and if k is negative, the graph is shifted down.

A negative value of h will
appear as an addition
symbol in the vertex form
of the equation, because
subtraction of a negative
is an addition of a positive.
For example, the function
f(x) = (x + 3)² written
in vertex form is
f(x) = (x − (−3))², so
the value of h is −3.

For any kind of function, the graph of f(x) + k is the graph of f(x) translated vertically—up if k is positive and down if k is negative—because the function value is adjusted at every point by exactly k units.

When a graph is shifted
without changing the
shape, orientation, or
magnitude of the graph, we
say that it is translated.

For example, the function f(x) = x⁴ − 2x² + 10 will always be exactly 10 units greater than the function f(x) = x⁴ − 2x² for any given x-value, so its graph is identical but shifted 10 units up.

The graph of f(x − h), for any kind of function, is the graph of f(x) translated h units to the right.

If you subtract a constant, h, from x, then the function value is what it would be for the x-value that is h units to the left in the parent function. In other words, the graph is shifted to the right when the subtracted h is positive. For example,g(x) = (x − 6)⁴ is f(x) = x⁴ shifted 6 units to the right. The value of g(8) is the same as the value of f(2), 16, as shown in the graphs on the following page.

The graph of f(x) = x³ is shown in the standard (x, y) coordinate plane below. Write an equation that describes the graph of the cubic function shifted 2 units to the right and 5 units down.

The function graph will be shifted 2 units to the right, so we must subtract 2 from x inside the cubed expression. The function graph will be shifted 5 units down, so we must subtract 5 from the function value (outside of the cubed expression). The translated function has the equation f(x) = (x − 2)³ − 5.

To check our answer, let’s test the equation in relation to the graph. The new function equation should shift each point 2 units right and 5 units down. Let’s sketch the translated graph next to the graph of f(x) = x³.

When x = 2, the original function f(x) = x³ has a value of 8, so the graph of f(x) = x³ passes through the point (2, 8). Let’s test the equation f(x) = (x − 2)³ − 5 to see if it produces the point that is on the translated function graph, 2 units to the right and 5 units below (2, 8), which is (4, 3). Test x = 4 in f(x) = (x − 2)³ − 5.

f(4) = (4 − 2)³ − 5 = 2³ − 5 = 8 − 5 = 3

The function correctly produced the point (4, 3). If you choose any point on f(x) = x³, the function f(x) = (x − 2)³ − 5 produces a point 2 units right and 5 units down from there.

Now you know how function graphs of degree n are translated when written in the form y = (x − h)n + k, where h and k are constants. However, you do not need to write a polynomial in this form to graph it. Solving for the zeros of a polynomial function allows you to make a rough sketch of its graph, using its x-intercepts and end behavior.

The roots of a single-variable polynomial expression are the solutions for the variable when the polynomial is set equal to 0. These roots are called zeros of the corresponding polynomial function. The zeros are the x-intercepts of the graph of the function.

In Example 13 (the ball-throw example), the roots of the quadratic are −1 and 4, because these are the values of t that are solutions to −16t² + 48t + 64 = 0. The values −1 and 4 are the zeros of the function h(t) = −16t² + 48t + 64. The graph of the function crosses the x-axis at −1 and 4.

The end behavior of a function describes how it behaves for very small (approaching negative infinity) and very large (approaching infinity) values of x. Because of the relationship of exponents, the value of the term with the greatest exponent overpowers any other terms within the expression, regardless of their coefficients. For example, for very large or very small values of x, y = x⁴ − 100x³ will have a greater x⁴ value than a 100x³ value, and the ends of the graph will follow the rules for x⁴.

When a polynomial function is of an even degree, the x-values at either end raised to that power produce very large positive numbers. When the leading coefficient for a polynomial function of even degree is positive, then both ends have very large positive values, and both arms of the graph point up. When the leading coefficient for a polynomial function of even degree is negative, this negative number multiplies by the large positive value of x raised to that power to create a negative value approaching negative infinity, and both arms of the graph point down.

When a polynomial function is of an odd degree, such as with a leading term of x³ or x⁵, then the sign of x is preserved for extreme values of x (a large negative x-value raised to an odd power produces a very large negative function value), and the arms of the function point in opposite directions from one another.

End Behavior of Polynomial Functions

A polynomial function of even degree with a positive leading coefficient has both arms pointing up.

A polynomial function of even degree with a negative leading coefficient has both arms pointing down.

A polynomial function of odd degree with a positive leading coefficient has its left arm pointing down and its right arm pointing up.

A polynomial function of odd degree with a negative leading coefficient has its left arm pointing up and its right arm pointing down.

Sketch the function y = −3x³ − 9x² + 30x.

First, factor the polynomial −3x³ − 9x² + 30x, set equal to 0.

Solve for each factor set equal to 0.

The zeros are −5, 0, and 2, so these are the x-intercepts of the function. The function is a cubic with a negative coefficient of x³, so the left end of the graph points up and the right end points down. Let’s sketch the graph using this information.

To find more points on
the graph, you would
use the given function
equation to solve for
y-values produced by
various x-values.

This is just a rough sketch. To graph more precisely, we would need to find at least a few more points on the graph.

Here is how you may see equations of polynomial graphs on the SAT.

The graph of the function f(x) contains the points (−6, 0), (3/2, 0), and (1, 0). Which of the following could be f?

A) f(x) = 2x³ + 7x² − 27x + 18

B) f(x) = 2x³ + 13x² + 3x − 18

C) f(x) = 2x³ − 17x² + 33x − 18

D) f(x) = 2x³ − 7x² − 27x − 18

Notice that the graph in Example 15 has two turning points, one of which is a local minimum and the other of which is a local maximum. A cubic may have two turning points, as in this case. It also may have no turning points, as in the graph of y = x³, shown in Example 14. A quadratic has exactly one turning point, which we call the vertex, as you learned in Algebra I. Straight lines have no turning points, as they never turn but continue forever in one direction.

A polynomial of degree n has at most (n − 1) turning points, which are local maximums or minimums.

With an understanding of how polynomial functions are graphed, we can also work in the other direction. Given the zeros of a polynomial function and a little more information about its magnitude, we can write the equation of the function.

What polynomial function is shown on the graph below?

The factor [x − (−2)] is
the same as (x + 2).

The three x-intercepts of this graph are −2, 1, and 4, so these are the zeros of the function. We can write each factor in the form (x − a), where a is a zero of the function. So, (x + 2), (x − 1), and (x − 4) are all factors of the polynomial. Let’s try multiplying them together.

(x + 2)(x − 1)(x − 4)

(x + 2)(x² − 5x + 4)

x³ − 3x² − 6x + 8

Both ends of the graph point upward, so it must be a polynomial function of an even degree. However, the polynomial x³ − 3x² − 6x + 8 is of degree 3—an odd degree, not even. How can that be? Well, when an x-intercept is also a local vertex, or turning point, then the zero is a repeated zero. In this case, −2 is a repeated zero, so we must use it twice when writing the polynomial.

If the polynomial function graphed were y = x⁴ − x³ − 12x² − 4x + 16, then it would have a y-intercept of 16 (the solution for y when x = 0). Instead, this graph shows a y-intercept of 4. If we multiply our polynomial by 1/4, that will adjust the magnitude, without affecting the zeros, to produce the correct y-intercept.

y = 1/4 (x⁴ − x³ − 12x² − 4x + 16)
y = 1/4 x⁴ − 1/4 x³ − 3x² − x + 4

The polynomial function shown in the graph is y = 1/4 x⁴ − 1/4 x³ − 3x² − x + 4 in standard form, or y = 1/4 (x + 2)(x + 2)(x − 1)(x − 4) in factored form.

Some polynomials have
fewer real zeros than
their highest degree
for other reasons. If the
graph reaches a turning
point before intersecting
with the x-axis (a local
maximum below the
x-axis or a local minimum
above the x-axis), then
the polynomial will have
two fewer real zeros.

The three given points all have an f(x)-value of 0, which means that they all represent x-intercepts of the function. So, the zeros of f(x) are −6, 3/2, and 1. We can write each factor of f(x) in the form (x − a), where a is a zero of the function.

(x − (−6))(x − 3/2)(x − 1)
(x + 6)(x − 3/2)(x − 1)

However, the multiplication of these factors will be a little messy with a fraction involved. Let’s write a different binomial factor to represent the zero 3/2. The only rule is that the binomial factor, when set equal to 0, must solve as x = 3/2. The binomial (2x − 3) fits this description.

A function with zeros −6, 3/2, and 1 could have other zeros, as well (which would result in additional binomial factors), or a different magnitude (which would mean multiplying this polynomial by a constant). However, we do not have to worry about these possibilities, because f(x) = 2x³ + 7x² − 27x + 18 is one of the given answer choices. The correct answer is (A).