could you help me with the following,I don't understand.
Find the x-intercepts of the polynomial function. State whether the graph crosses the x-axis, or touches the x-axis and turns around, at each intercept.
1.f(x) = (x + 1)(x - 3)(x - 1)^2
5.f(x) = x4 - 16x2
7.x4 + 5x3 - 84x2 = 0
5.f(x) = x^4 - 16x^2
7.x^4 + 5x^3 - 84x^2 = 0
4th degree functions can have a maximum of 4 x-intercepts (in general a nth degree function can have at most n x-intercepts)
They tend to look like a W
for 1.
the factors are (x+1)(x-3)(x-1)^2
so there are x-intercepts at -1 and 3.
The factor (x-1)^2 or (x-1)(x-1) tells us that the graph "touches" the x-axis at 1
Whenever you have a squared factor such as (x-a)^2 the curve would touch at a
For 7. factor it this way
x^2(x^2 + 5x - 84)
=x^2(x+12)(x-7)
the graph would touch at the origin, and cut the x-axis at -12 and 7
let me know what you got for #5
Of course! I can help you with finding the x-intercepts of polynomial functions and determining their behavior on the graph. Let's go through each polynomial one by one:
1. f(x) = (x + 1)(x - 3)(x - 1)^2
To find the x-intercepts, we need to set f(x) equal to zero and solve for x:
0 = (x + 1)(x - 3)(x - 1)^2
To determine whether the graph crosses or touches the x-axis, we will analyze the multiplicity of each factor:
- The factor (x + 1) has an odd multiplicity of 1, so the graph crosses the x-axis at x = -1.
- The factor (x - 3) has an odd multiplicity of 1, so the graph crosses the x-axis at x = 3.
- The factor (x - 1) has an even multiplicity of 2, so the graph touches the x-axis at x = 1 and turns around.
Therefore, the x-intercepts of f(x) = (x + 1)(x - 3)(x - 1)^2 are -1 (crosses), 1 (touches and turns around), and 3 (crosses).
2. f(x) = x^4 - 16x^2
To find the x-intercepts, we set f(x) equal to zero and solve for x:
0 = x^4 - 16x^2
This equation is in quadratic form, so we can factor it:
0 = x^2(x^2 - 16)
Now, we have two cases to consider:
Case 1: x^2 = 0
In this case, x = 0, which is an x-intercept.
Case 2: x^2 - 16 = 0
This quadratic equation can be factored further:
0 = (x - 4)(x + 4)
So, x = 4 and x = -4 are also x-intercepts.
Therefore, the x-intercepts of f(x) = x^4 - 16x^2 are -4, 0, and 4.
3. f(x) = x^4 + 5x^3 - 84x^2 = 0
To find the x-intercepts, we set f(x) equal to zero and solve for x:
0 = x^4 + 5x^3 - 84x^2
This equation cannot be factored easily, so we will use another method. We can use a numeric method like graphing or numerical approximation to find the x-intercepts.
Using a graphing calculator or software, we find that the x-intercepts are approximately x = -6.739, x = -3.714, x = 0, and x = 12.453.
Therefore, the x-intercepts of f(x) = x^4 + 5x^3 - 84x^2 are approximately -6.739, -3.714, 0, and 12.453.
Let me know if you need any further assistance!
Of course! I'd be happy to help you with your questions.
1. To find the x-intercepts of the polynomial function f(x) = (x + 1)(x - 3)(x - 1)^2, you need to set f(x) equal to zero and solve for x. So, you have the equation (x + 1)(x - 3)(x - 1)^2 = 0.
Setting each factor equal to zero gives you three possible x-intercepts:
x + 1 = 0, which gives x = -1 as one x-intercept.
x - 3 = 0, which gives x = 3 as another x-intercept.
(x - 1)^2 = 0, which gives x = 1 as a double x-intercept.
To determine whether the graph crosses the x-axis or touches the x-axis and turns around at each intercept, you can look at the multiplicities of each factor.
For x = -1, the factor (x + 1) has a multiplicity of 1, meaning the graph crosses the x-axis at x = -1.
For x = 3, the factor (x - 3) has a multiplicity of 1, so the graph also crosses the x-axis at x = 3.
For x = 1, the factor (x - 1) has a multiplicity of 2. This means the graph touches the x-axis at x = 1 and turns around without crossing it.
2. For the polynomial function f(x) = x^4 - 16x^2, you also need to set f(x) equal to zero and solve for x. The equation is x^4 - 16x^2 = 0.
Factoring out common terms, you get x^2(x^2 - 16) = 0.
Setting each factor equal to zero gives you two possible x-intercepts: x^2 = 0, which gives x = 0, and x^2 - 16 = 0, which gives x = ±√16 = ±4.
To determine how the graph behaves at each intercept, you can consider the multiplicities of the factors.
For x = 0, the factor x^2 has a multiplicity of 2. This means the graph touches the x-axis at x = 0 and turns around without crossing it.
For x = 4 and x = -4, the factor x^2 - 16 has a multiplicity of 1. So, the graph crosses the x-axis at these intercepts.
3. For the equation x^4 + 5x^3 - 84x^2 = 0, you can factor out the common term x^2 to get x^2(x^2 + 5x - 84) = 0.
Setting each factor equal to zero gives you two possible solutions: x^2 = 0, which gives x = 0, and x^2 + 5x - 84 = 0.
To solve the quadratic equation x^2 + 5x - 84 = 0, you can factor it as (x + 12)(x - 7) = 0, which gives x = -12 and x = 7 as additional solutions.
For x = 0, the factor x^2 has a multiplicity of 2, so the graph touches the x-axis at x = 0 and turns around without crossing it.
For x = -12 and x = 7, the factor x^2 + 5x - 84 has a multiplicity of 1 each, so the graph crosses the x-axis at these intercepts.
I hope this explanation helps you understand how to find x-intercepts and determine the behavior of the graph at each intercept. Let me know if you have any further questions!