Jan is asked to sketch a graph of f(x) = −2x^2(x−4)^2(x+1)^3 . How many turning points should her graph include?(1 point)
1. 5 at most
2. 2 at most
3. 6 at most
4. 7 at most
The given function is a polynomial of degree 7, so it can have at most 7 turning points. Therefore, the correct answer is 4. 7 at most.
To determine the number of turning points for the graph of f(x) = −2x^2(x−4)^2(x+1)^3, we need to consider the multiplicity of each factor.
The multiplicity of a factor determines how many times the graph touches or crosses the x-axis.
In this case, the factor (x−4) has a multiplicity of 2, which means this graph will touch or cross the x-axis at x=4 twice. Similarly, the factor (x+1) has a multiplicity of 3, so the graph will touch or cross the x-axis at x=-1 three times.
Since the graph is a polynomial of degree 8 (2 from each factor), the total number of turning points on the graph will be 8 - 1 (one less than the degree), giving us a total of 7 turning points.
Therefore, the correct answer is 4. 7 at most.
To determine the number of turning points on the graph of the function f(x) = −2x^2(x−4)^2(x+1)^3, we need to analyze the behavior of the function and its derivatives.
To find the turning points, we first need to find the critical points by setting the derivative of the function equal to zero.
1. Find the derivative of f(x) with respect to x:
f'(x) = -4x(x−4)^2(x+1)^3 - 2x^2(2)(x−4)(x+1)^3 - 3(2)(x−4)^2(x+1)^2
2. Set f'(x) = 0 and solve for x:
-4x(x−4)^2(x+1)^3 - 4x^2(x−4)(x+1)^3 - 6(x−4)^2(x+1)^2 = 0
This equation might be complicated to solve by hand. However, we can use a graphing calculator or software to find the critical points.
3. Once you find the critical points, they represent the potential turning points on the graph of f(x).
Since solving this equation can be complex and time-consuming, it is more practical to use graphing software or a graphing calculator to determine the number of turning points.
Based on the multiple-choice options given, the correct answer is:
3. 6 at most