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the y intercept occurs when x = 0 so what do you think the y-intercepts is?
how would I find the Y intercept?
plug in x = 0 and work it out
0^3 - 18(0)^2 + 101(0) - 180 = ?
so x=0 and y=-180 how would I figure out the shape of the graph?
well try and find the x intercepts first they occur when y = 0 so we have the equation x^3 − 18x^2 + 101x − 180 = 0
so far I got 180=x^3 - 18^2 + 101x
- solve for x by the rational roots theorem the roots could be factors 0f 180 like +/- 1, +/- 3, +/- 6 , +/-5,+/- 6, +/- 9 so there's a lot to try out!
or we might be able to factor by grouping which would be a lot quicker:- x3 − 18x2 + 101x − 180
I got 3 answers and I don't know which one is right x = 9 x = 5 x = 4
nope cant do it that way
how did you get those?
- they might all be right. A cubic equation will have 3 roots.
f(5) = 5^3 - 18(25) + 101(5) - 180 = 0 yes 5 is a root
im so confused, so x would be 5 and y is -180
did you get a calculator or wolfram to do it?
no x = 5 when y = 0
how did you get those 3 roots They are all correct I just checked them on my calc.
I couldn't figure it out so I used a calculator
oh i see well another way is to to try ouy values such as 3, 5 etc and once you have found one - say 5- you divide the function by x - 5 using long or synthetic division to get a quadratic which you solve to get the other 2.
to find the shape you can differentiate and find slopes at different values of x We know it is probably a double loop because there are 3 rational roots.
so the shape is a loop?
either like that
or like this |dw:1441220677250:dw|
f(3) = -12 so it will be the first one
wow thank you so much!
guess you should include the y intercept |dw:1441220974136:dw|
- in fact i need not have found f(3) because the y intercept was negative anyway
- it had to be the first shape not the second. - i'd forgotten about the y-intercept