Day 9: Evaluating Polynomial Functions
The value of a polynomial function can be determined in 3 different ways.
1. If the graph is given, the coordinates of a point can be approximated by looking at the graph.
2. The function can be graphed using technology (graphing calculator) and the calculator can be
used to determine the y-value for a given x-value.
3. The given x-value can be substituted into the equation for the polynomial function and the
expression can then be evaluated (following order of operations) to determine the corresponding
y-value.
Example 1:
Given the following graph of y = f(x), determine
a) the value of y when x = 2.
b) the value of x when y = 8.
Example 2:
Use the equation of the function y = x3 – x2 + x, to determine
a) the value of y when x = 3.
b) the value of x when y = 0.
Example 3:
Use a graphing calculator and the equation y = 2x3 – x2 + 7 to determine
a) the value of x when y = 19. (use the TBLSET feature)
b) the value of y when x = 4. (use the value feature)
c) the value of x when y = -3. (use the intersection of graphs method)
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Function Notation:
The notation y = f(x) is called function notation. The ‘f’ signifies that the graph of y = f(x) will be a
function (It will pass a vertical line test.).
For a given value x = a, the value f(a) is the y-value for the function when x = a and can be represented
by the ordered pair (a, f(a) ) on the graph of the given function.
The value f(0) for a polynomial function represents the value of the y-intercept of the polynomial
function, f(x).
Example 4:
a) x = 2
Consider the polynomial function f(x) = 2x4 – x3 + 4x2 – 5x + 7. Evaluate the polynomial
function for each of the following values.
b) x = -3
c) x = 0
Homework:
1. Given the following graph of y = f(x), determine
a) the value of y when x = 2.
b) the value of x when y = 5.
c) f(0)
2. Use the equation of the function
f(x) = -2x3 – 5x2 + x - 3, to determine
a) the value of y when x = 3.
b) f(-1)
c) f(0)
3. How could you determine the value of x when y = 7 for the function f(x) = -2x3 – 5x2 + x – 3?
4. Consider the polynomial function f(x) = 2x3 + 7x2 + 3x - 4. Evaluate the polynomial function for
each of the following values.
a) x = 4
b) x = -3
c) x = 1
5. Evaluate f(7) for the function f(x) = 4x – 1 . Does f(x) represent a polynomial function? Explain.
6. Given the polynomial function f(x) = x2 + 2x, determine an expression for f(a).
7. Time permitting, check your answers to #2, 3, and 4 using a graphing calculator.
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Calculator Instructions:
1. Calculating a value using the graphing calculator.
For the function, f(x) = -3x + 17, find f(3). To find f(3), Press
, and enter -3x + 17 for . Make sure
that the given x-value is shown on the calculator screen. If this is not the case, adjust the window to
include the x-value at which the function will be evaluated. In our case, we need to evaluate the
function at x = 3 so be sure that the graph includes the x-value of 3.
Next press 2nd TRACE[Calc] to get the calculate menu. Press ENTER to accept the selection of
1:value. Type 3 next to the X= at the bottom of the screen and then press ENTER.
A Y=8 appears at the bottom of the screen. That means, for this function, when x = 3, y = 8.
Two other ways to say that are that (3, 8) is a point of this function and we can also say f(3 )= 8.
2. Determining values using the table.
To view a table with values that your equation produces, enter 2nd TBLSET. You can change the starting
value in the table (TblStart”) or the increment between X values in the table (“
value at these locations on the screen. Generally, we keep the “AUTO” options selected in order to get
an automatic table of X and corresponding Y values. The following setup will produce a table with 0 at
shown. Note: You can “scroll” up and down in the “X” column as far as you’d
like.
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3. Determining a y-value using the Intersection of Graphs Method
Any type of equation can be solved by this method. For example, solve the equation
x 2 + x = 20 using this method.
.
This method uses graphing of functions to solve an equation.
STEP 1: Set
equation.
STEP 2: Graph
equal to the left side of the equation and
and
equal to the right side of the
in the same viewing rectangle.
STEP 3: Locate any points of intersection. The set of x-values of these points of intersection
corresponds to the solution set of the equation. Some solutions may be exact, while others may
be only approximate solutions.
STEP 1: Press the
the equation for .
button. Then enter the left side of the equation for
and the right side of
STEP 2: Since
has a graph which is a horizontal line at 20 on the y-axis, we will set our
initial window as follows: [-10,10,1] by [0,30,2]. This way we are assured that the horizontal
line, y = 20, will be visible. After the window is set, press GRAPH. Two intersection points are
visible in the screen below.
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STEP 3: Now press the 2nd key, then TRACE [Calc], then select 5:intersection. When First
curve? appears, press ENTER. When Second curve? appears press ENTER again.
Since there are two intersection points, when Guess? appears, move the curser close to one of
the points using the ARROW keys and then press ENTER. The screens below show finding the
leftmost intersection first.
The final screen above shows that the intersection point is (-5,20). If there are more points of
intersection, move the cursor near each one and repeat the process. The X value of the
intersection point, X = -5, is one of the two solutions to the equation. Next repeat the entire
process, with the exception at GUESS?, move the curser to the rightmost intersection, then press
ENTER. The results are shown below.
The final screen above shows that the intersection point is (4,20). If there are more points of
intersection, move the cursor near each one and repeat the process. The X value of the
intersection point, X = 4, is one of the two solutions to the equation.
Thus the solutions to the equation, x 2 + x = 20, is x = - 5 or x = 4.
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