Practice test #10 and #13
For number ten you can see from the graph that {-3, -1, -1, 1, 3, 3} are the roots. I started by turning the roots into factors so I had (x+3)*(x+1)^2*(x-1)*(x-3)^2. From there I'm not sure whether or not to add or subtract anything since the graph has a y intercept not at zero. Also, how would you come up with a second equation for this graph?
For number 13 I don't really know where to start. a≠1 is a solution how would you use that information to start a proof?
For number 13 I don't really know where to start. a≠1 is a solution how would you use that information to start a proof?
posted by Carrie Desmond at 11:57 AM
6 Comments:
You're right about the roots and factors (i actually hadn't thought to square (x+1) and (x-3) until you reminded me... thanks for that). You can multiply all six factors by a scale factor called a. The graph includes the point (-2, -30), so you can plug these values in for x and y in the equation y=a*(x+3)*(x+1)^2*(x-1)*(x-3)^2 to find a. (I got a value of 6/15 for a, but you might want to check my algebra.) So that's one equation. For my second equation, I took (x+3) and one of the (x-3)s and made them (x^2-9). I similarly took (x-1) and one of the (x+1)s and made them (x^2-1). This gives a new and different equation of y=(6/15)*(x^2-9)*(x^2-1)*(x-3)*(x+1).
For #10, what i did was, from the graph {-3,-1,1,-3} are the roots--my polynomial was a four-degree polynomial, so i got y=a*(x+3)(x+1)(x-3)(x-1), then i substituted in (-2,-30), so a=2. My first equation was y=2(x+3)(x+1)(x-1)(x-3). Since (a+b)(a-b)=a^2-b^2, then my second equation was y=2(x^2-9)(x^2-1).
For #13, i'm not sure of how to start, but i think you may need to use the fact, which states sum of the roots=sum of the real roots+sum of the complex roots, and the sum of the roots is 0. i'm really not sure of this question.
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Simply multiply the entire equation by some coefficent to get another equation this however will not cange the y-intercept to change the y-intercept you have to add or subtract some number until you get the same equation.
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