Question 1058117: Hi!
Can you please help me solve this question.
Its using a graph I am positive about my equation.
Based on my equation- y=1(x+2)^2(x-2)
My question is State the factors based on the graph include multiplicities
2nd question is State a point on the graph(different than the zeros). Substitute the x and y values into your factored equation and solve for a.
3rd question: Multiply your factors to find the standard form of polynomial equation
4rth question: please state the maximum and minimum , the intervals where the graph is increasing and decreasing.
That's all I need help with
I would really appreciate it if you can solve and explain it correctly
Thank You so much
sincerely
Bella
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! I will assume f(x)=1(x+2)^2*(x-2)
(x+2) with multiplicity 2 (it will bounce on the x-axis)
The roots are -2 with multiplicity 2 and 2.
The polynomial is x^3+2x^2-4x-8 as I have written the above. I don't know if the 1 was intended, but I can't solve for a if I don't know one point on the graph, so I will continue with this. The domain is all x and the range is the same. If I substitute a point other than the roots, like 0, f(0)=-8; (0,-8) is a point.
The derivative is 3x^2+4x-4
Set that equal to 0
(3x-2)(x+2)=0
x=(2/3) and x=-2. Those are critical values. f(-2)=0; f(2/3)=-256/27
To the left of -2 the function is negative, and to the right of -2 is negative, so -2 is a maximum.
Pick x=1/2, and (5/2)^2*(-3/2) is -9 3/8, which is slightly larger than -256/27. f(1)=-9, which is larger than -256/27, so x=2/3 is a local minimum.
Second derivative is 6x+4. It is positive when x is greater than -2/3 and negative when x is less than -2/3, so it is convex up for the x=-2 (negative value) and convex down for x=2/3 (positive value)
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