# quartic graph help

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#1
Ok so I get how they found the y intercept, the roots (2 roots as the quadratics discriminant is negative) but how do they tell where the graphs gradient changes (I'm talking about the little concave, little change in gradient from -5 to -280). I did this question and I drew it out with the concave (change in gradient) not between -5 to -280 but AFTER x=0. So how do they know where it changes? For previous ones, I input values to work out the quartic graph (x = 1 x= 2 etc. to tell whether y would be greater or lesser, telling me if the gradient changes) but for this one it seems that the gradient from -5 to -280 always stays negative, which doesn't help much.

thanks for any help
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4 days ago
#2
(Original post by pondering-soul)
Ok so I get how they found the y intercept, the roots (2 roots as the quadratics discriminant is negative) but how do they tell where the graphs gradient changes (I'm talking about the little concave, little change in gradient from -5 to -280). I did this question and I drew it out with the concave (change in gradient) not between -5 to -280 but AFTER x=0. So how do they know where it changes? For previous ones, I input values to work out the quartic graph (x = 1 x= 2 etc. to tell whether y would be greater or lesser, telling me if the gradient changes) but for this one it seems that the gradient from -5 to -280 always stays negative, which doesn't help much.

thanks for any help
If you complete the square on the quadratic factor what do you get?
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#3
(Original post by mqb2766)
If you complete the square on the quadratic factor what do you get?
so I did it, got the answer in the attached file. I guess what your trying to say is that its gradient changes at 5/2, given the info that I just got from completing the square. 31/4 would have been the y intercept if it was only a quadratic but since its a quartic graph that's not the case. However this only leaves me with a lot more questions... how do we know the the values for the completed square quadratic don't translate to the OTHER turning point, since there are 2 quadratics and 2 turning points (well, one turning point and one wierd thingie). (what I'm trying to say is I saw (x + 5/2)^2 + 31/4, so given this info we can assume that gradient changes at 5/2 which the graph DOES show, but why is this change because of the (x+ 5/2 )^2 + 31/4 quadratic and not the other one?)

thanks for the helps regardless... it's a long one
Last edited by pondering-soul; 4 days ago
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4 days ago
#4
(Original post by pondering-soul)
so I did it, got the answer in the attached file. I guess what your trying to say is that its gradient changes at 5/2, given the info that I just got from completing the square. 31/4 would have been the y intercept if it was only a quadratic but since its a quartic graph that's not the case. However this only leaves me with a lot more questions... how do we know the the values for the completed square quadratic don't translate to the OTHER turning point, since there are 2 quadratics and 2 turning points (well, one turning point and one wierd thingie). (what I'm trying to say is I saw (x + 5/2)^2 + 31/4, so given this info we can assume that gradient changes at 5/2 which the graph DOES show, but why is this change because of the (x+ 5/2 )^2 + 31/4 quadratic and not the other one?)

thanks for the helps regardless... it's a long one
The gradient changes at -5/2. This is why the link occurs around that value.
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4 days ago
#5
(Original post by pondering-soul)
so I did it, got the answer in the attached file. I guess what your trying to say is that its gradient changes at 5/2, given the info that I just got from completing the square. 31/4 would have been the y intercept if it was only a quadratic but since its a quartic graph that's not the case. However this only leaves me with a lot more questions... how do we know the the values for the completed square quadratic don't translate to the OTHER turning point, since there are 2 quadratics and 2 turning points (well, one turning point and one wierd thingie). (what I'm trying to say is I saw (x + 5/2)^2 + 31/4, so given this info we can assume that gradient changes at 5/2 which the graph DOES show, but why is this change because of the (x+ 5/2 )^2 + 31/4 quadratic and not the other one?)

thanks for the helps regardless... it's a long one
I'm not sure what either of you means when you say that the gradient changes at -5/2. The gradient changes everywhere. To get a better sketch you will either have to differentiate, twice if you want to find the points of inflection, or do a table of values. The working provided is not sufficient to produce the given answer, in my opinion.
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#6
(Original post by Plücker)
I'm not sure what either of you means when you say that the gradient changes at -5/2. The gradient changes everywhere. To get a better sketch you will either have to differentiate, twice if you want to find the points of inflection, or do a table of values. The working provided is not sufficient to produce the given answer, in my opinion.
oh right yeah I didn't mean the gradient I meant like... that little point at -5/2 where the curve kind of changes steeply, I guess it would be the turning point if it was a quadratic but the line doesn't turn upwards it just keeps going down). Also, that's what I kind of thought, the working given in mark scheme doesn't really show how it got the steep change (I mean at x = -5/2 you can see the line kind of drops)
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#7
(Original post by pondering-soul)
oh right yeah I didn't mean the gradient I meant like... that little point at -5/2 where the curve kind of changes steeply, I guess it would be the turning point if it was a quadratic but the line doesn't turn upwards it just keeps going down). Also, that's what I kind of thought, the working given in mark scheme doesn't really show how it got the steep change (I mean at x = -5/2 you can see the line kind of drops)
I think perhaps I would get the full marks for my attempt because the question / mark scheme may not really care too much about where this inflection (idk what that is I guess thats what u call it) occurs so long as the roots and y intercept is ok
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4 days ago
#8
(Original post by Plücker)
I'm not sure what either of you means when you say that the gradient changes at -5/2. The gradient changes everywhere. To get a better sketch you will either have to differentiate, twice if you want to find the points of inflection, or do a table of values. The working provided is not sufficient to produce the given answer, in my opinion.
1
4 days ago
#9
(Original post by pondering-soul)
I think perhaps I would get the full marks for my attempt because the question / mark scheme may not really care too much about where this inflection (idk what that is I guess thats what u call it) occurs so long as the roots and y intercept is ok
Just in case they expect more you can quite easily do a table of values using the table mode on your calculator. You could of course plot it on a graphical calculator.
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4 days ago
#10
(Original post by mqb2766)
Is that useful, given that it can coincide with a turning point, a point of inflection or a point of no particular interest?
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4 days ago
#11
(Original post by Plücker)
Is that useful, given that it can coincide with a turning point, a point of inflection or a point of no particular interest?
,Sure it's not a general result and I suspect the fine detail may be beyond what is required. The [-5,4] gives us the interval of interest as outside this interval, the 3 factors behave the same. Inside the interval the two quadratic factors are
p(x) = x^2 + x - 20
q(x) = x^2 + 5x + 14
Completing the square on each gives minimum at (-1/2, -20) and (-5/2,8) roughly. As the derivative of the quartic is
p'q + pq'
We need a bit of reasoning about the signs of the factors and the derivatives to approximate what will happen.

In (-5,-5/2) the second term of the derivative is going to zero because of q'. The first term is negative so the quartic is decreasing.
In (-5/2,-1/2) both terms are negative as q and q' are negative and p and p' are positive.
in (-1/2,4) the quartic derivative goes from negative to positive, which corresponds to the single turning point somewhere a bit to the right of -1/2.
Last edited by mqb2766; 4 days ago
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4 days ago
#12
(Original post by mqb2766)
,Sure it's not a general result and I suspect the fine detail may be beyond what is required. The [-5,4] gives us the interval of interest as outside this interval, the 3 factors behave the same. Inside the interval the two quadratic factors are
p(x) = x^2 + x - 20
q(x) = x^2 + 5x + 14
Completing the square on each gives minimum at (-1/2, -20) and (-5/2,8) roughly. As the derivative of the quartic is
p'q + pq'
We need a bit of reasoning about the signs of the factors and the derivatives to approximate what will happen.

In (-5,-5/2) the second term of the derivative is going to zero because of q'. The first term is negative so the quartic is decreasing.
In (-5/2,-1/2) both terms are negative as q and q' are negative and p and p' are positive.
in (-1/2,4) the quartic derivative goes from negative to positive, which corresponds to the single turning point somewhere a bit to the right of -1/2.
I see. Good stuff. I'll probably just recommend that people find any intersections with the axes and then use the table mode to fill in the gaps a little.
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4 days ago
#13
(Original post by Plücker)
I see. Good stuff. I'll probably just recommend that people find any intersections with the axes and then use the table mode to fill in the gaps a little.
I'd have to generally agree apart from a couple of minor things
* it shows that all the action happens in the interval (-5,4) as this may not be obvious and gives bounds for table entries
* If it was an entrance exam type question (it isn't) where they seem to like completing the square and require a bit of deeper reasoning about how common results (quadratics) can be applied in novel situations.
Last edited by mqb2766; 4 days ago
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