Graph Polynomials without a Calculator
Graph Polynomials Without a Calculator- How it Works - Video
Intermediate Value Theorem (IVT)
Intermediate Value Theorem (IVT):
Here we have two models. The models are the same except the one on the left is decreasing and the one on the right is increasing.
No matter, which way our graph is going, we will have the same scenario. Remember IVT states there is always a number in between two numbers as long as the two numbers are not equal.
Mathematical that is if f(a) ≠ f(b) for a < b then f takes a value between f(a) and f(b).
We can use this theorem to tells us where the zeros (x-intercepts - roots - answers) are. Remember 0 is neither positive nor negative. If f(a) = positive and f(b) = negative like in the graph on the left, we will have a zero somewhere in between, in this case at z. If f(a) = negative and f(b) = positive like in the graph on the right, we still will have a zero somewhere in between, in this case at z.
We can use this theorem to find zeros and sketch polynomials.
Example 1
Example 1:
We have the function f(x) = x3 + x2 - 4x - 4. The first step is find all the zeros. If we don't have a calculator, one way to find our zeros is to factor by grouping.
f(x) = x3 + x2 - 4x - 4
f(x) = x2 * (x + 1) - 4 * (x + 1)
f(x) = (x2 - 4)(x + 1)
f(x) = (x - 2)(x + 2)(x + 1)
Factored out x2 in the first group and factored -4 in the second group.
Combined x2 and 4.
Difference of squares.
After setting (x -2) = 0, (x + 2) = 0, and (x + 1) = 0, we find our zeros, x = -2, -1, 2. So we plot those points, (-2, 0), (-1, 0), and (2, 0) on our graph.
To start our sketch, we draw vertical lines through our zeros to create sections. To find which our graph is, we can substitute any input value into factored function. You can use the original function, but it will be harder to tell if it positive or negative because of the different exponents and coefficients.
Once you have found if the output value is positive or negative, you can put a check mark there. No matter what point you pick in that section, will the same as in positive or negative.
Section 1
f(x) = (x - 2)(x + 2)(x + 1)
f(-9) = ([-9] - 2)([-9] + 2)([-9] + 1)
f(-9) = (-)(-)(-)
f(-9) = negative
Substituted -9.
Found if negative or positive.
3 negatives = negative.
Section 2
f(x) = (x - 2)(x + 2)(x + 1)
f(-1.5) = ([-1.5]-2)([-1.5]+2)([-1.5]+1)
f(-9) = (-)(+)(-)
f(-9) = positive
Substituted -1.5.
Found if negative or positive.
2 negatives = positive.
Section 3
f(x) = (x - 2)(x + 2)(x + 1)
f(0) = ([0] - 2)([0] + 2)([0] + 1)
f(0) = (-)(-)(-)
f(0) = negative
Substituted 0.
Found if negative or positive.
1 negative = negative.
Section 4
f(x) = (x - 2)(x + 2)(x + 1)
f(10) = ([10] - 2)([10] + 2)([10] + 1)
f(10) = (+)(+)(+)
f(10) = positive
Substituted 0.
Found if negative or positive.
0 negatives = positive.
We put check marks where the result of each each section. These check marks tell us where our graph will be.
Example 2
Example 2:
We have the function f(x) = x4 - x3 - 6x2. The first step is find all the zeros. If we don't have a calculator, one way to find our zeros is to factor by grouping, but we don't have 4 terms. So let's check to see if each term shares a variable and/or a number.
f(x) = x4 - x3 - 6x2
f(x) = x2 * (x2 - x - 6)
f(x) = x2 * (x - 3)(x + 2)
Factored out x2 from each term.
Factored the trinomial parentheses.
After setting x2 = 0, (x - 3) = 0, and (x + 2) = 0, we find our zeros, x = -2, 0, 0, 3. So we plot those points, (-2, 0), (0, 0), and (3, 0) on our graph. We don't need to repeat the second zero at x2, but we will use the fact that our zero repeats with our graph.
To start our sketch, we draw vertical lines through our zeros to create sections. To find which our graph is, we can substitute any input value into factored function. You can use the original function, but it will be harder to tell if it positive or negative because of the different exponents and coefficients.
Once you have found if the output value is positive or negative, you can put a check mark there. No matter what point you pick in that section, will the same as in positive or negative.
Section 1
f(x) = x2 * (x - 3)(x + 2)
f(-10) = [-10]2 * ([-10] - 3)([-10] + 2)
f(-10) = (+)(-)(-)
f(-10) = negative
Substituted -10.
Found if negative or positive.
2 negatives = positive.
Section 2
f(x) = x2 * (x - 3)(x + 2)
f(-1) = [-1]2 * ([-1] - 3)([-1] + 2)
f(-1) = (+)(-)(+)
f(-1) = positive
Substituted -1.5.
Found if negative or positive.
2 negatives = positive.
Section 3
f(x) = x2 * (x - 3)(x + 2)
f(1) = [1]2 * ([1] - 3)([1] + 2)
f(1) = (+)(-)(+)
f(1) = negative
Substituted 1.
Found if negative or positive.
1 negative = negative.
Section 4
f(x) = x2 * (x - 3)(x + 2)
f(20) = [20]2 * ([20] - 3)([20] + 2)
f(20) = (+)(+)(+)
f(20) = positive
Substituted -1.5.
Found if negative or positive.
0 negatives = positive.
We put check marks where the result of each each section. These check marks tell us where our graph will be.
Here we have a two negatives in a row because we have a zero that repeats itself at x2 = 0. So it bounces back once it reaches (0, 0) instead of going through the x-axis.