Question 849554: I was given a graph. I was told to find the x-and y-intercepts, determine if leading coefficient is positive or negative, whether the degree of the polynomial function is odd or even, is the multiplicity of x=-4 odd or even? If (1,-5) is a point of the graph, (which it is), find the equation of the function.
x-ints are: (0,0), (2,0), (5,0), (-4,0), (-4,0)
y-int is: (0,0)
Multiplicity of -4 is even, because it touches doesn't cross the x-axis.
I have the degree as odd, because there are technically 5 x-ints.
I have the leading coefficient as negative. If you need a picture of the graph I am able to scan a copy of what I have if I get an e-mail address to send it to.
Some other points to help out are: (-2,10) and (4,25)
Thank you in advance.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you can find the equation from the roots.
the roots are:
x = 0
x = 2
x = 5
x = -4
x = -4
multiplying all those together and I got:
x^5 + x^4 - 30x^3 - 32x^2 + 160x
i then graphed that equation.
that graph looked like the picture, but was oriented in revedrse, so i multiplied it by -1 and got:
-x^5 + x^4 + 30x^3 + 32x^2 - 160x
that graph was oriented right but the point (1,-5) was not on the graph.
instead i got the point (1,-100).
this was much to low, so i divided the equation by 20 and got:
-.05x^5 + .05x^4 + 1.5x^3 + 1.6x^2 - 8x
that allowed the point (1,-5) to be on the graph and kept all the roots where they were supposed to be.
the graph looks like this:
it is a 5th degree equation.
the coefficient of the leading term had to be negative, so it start high on the left and ends low on the right.
the multiplicity at x = -4 is even (2 roots at x = -4).
the point (1,-5) is on the graph.
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