# quartic graph help

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

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

thanks for any help

0

reply

Report

#2

(Original post by

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

**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

0

reply

(Original post by

If you complete the square on the quadratic factor what do you get?

**mqb2766**)If you complete the square on the quadratic factor what do you get?

thanks for the helps regardless... it's a long one

Last edited by pondering-soul; 4 days ago

0

reply

Report

#4

(Original post by

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

**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

0

reply

Report

#5

**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

0

reply

(Original post by

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.

**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.

0

reply

(Original post by

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)

**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)

0

reply

Report

#8

**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

reply

Report

#9

(Original post by

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

**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

0

reply

Report

#10

(Original post by

The sign of the gradient of the quadratic factor.

**mqb2766**)The sign of the gradient of the quadratic factor.

0

reply

Report

#11

(Original post by

Is that useful, given that it can coincide with a turning point, a point of inflection or a point of no particular interest?

**Plücker**)Is that useful, given that it can coincide with a turning point, a point of inflection or a point of no particular interest?

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

0

reply

Report

#12

(Original post by

,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.

**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.

0

reply

Report

#13

(Original post by

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.

**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.

* 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

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top