geometrical applications (1 Viewer)

[maths lover]

prove that the graph of <text>y=ax^{3}+bx^{2}+cx+d has two distinct turning points if b^{2}> 3ac. hence find the values of a,b,c and d for which the graph formed has the turning points at (0.5,1) and (3/2,-1). thankyou.<text></text></text>

 

[OldMathsGuy]

Hint:
Differentiate the expression with respect to x (remember that you treat the other pronumerals as constants). This gives you a quadratic gradient function. From here you need to use your knowledge of roots and the discriminant to determine when there will be two solutions (rather than one or no solutions). From this you should be able to establish the inequality.
The second part will just be about constructing equations from that information and solving them (most likely simultaneously).

Does that help?
I prefer to give hints to begin with; you learn from the struggle of completing more so than that of reading and repeating.

Best Regards
OldMathsGuy

 

[D94]

As for the second part, remember how the two x coordinates are what you use to find the stationary points, and whether they are maximum or minimum. So construct an equation with the two x coordinates and equate the coefficients with the differentiated equation.

 

[maths lover]

Hint:
Differentiate the expression with respect to x (remember that you treat the other pronumerals as constants). This gives you a quadratic gradient function. From here you need to use your knowledge of roots and the discriminant to determine when there will be two solutions (rather than one or no solutions). From this you should be able to establish the inequality.
The second part will just be about constructing equations from that information and solving them (most likely simultaneously).

Does that help?
I prefer to give hints to begin with; you learn from the struggle of completing more so than that of reading and repeating.

Best Regards
OldMathsGuy

yes thanks that was helpful, kinda what i tried at first. i will give it another try and if i still cant get it post up again.

 

[maths lover]

repped both, i originally had the differentiated equation just didn't realize to use the discriminate seems kind of stupid now.

 

[nightweaver066]

lol if u get the 2nd part can you post up how to do it!

For the second part, you would need to use the first derivative:



As the question gave you the turning points, and these points are where the gradients are = 0, simply sub in the x values to obtain 2 equations.

Now you sub in the turning points in to the equation of the curve to obtain another 2 equations.

Using simultaneous equations, the answers will work out nicely to be:


.. if my working out was correct.

 

[maths lover]

lol if u get the 2nd part can you post up how to do it!

second one isnt actually that hard. all you have to do is basically use sum and product of roots knowing that roots of the differentiated equation represent the x-coord of the max/min turning point, get it?.

 

[nightweaver066]

second one isnt actually that hard. all you have to do is basically use sum and product of roots knowing that roots of the differentiated equation represent the x-coord of the max/min turning point, get it?.

You can't really do it that way as using the sum and product of roots would get you two equations, both of them have 1 variable in common where as the other variable is different meaning you cannot solve these using simultaneous equations.

Sum of roots (from first derivative):

Product of roots (from first derivative):


If you try solve these, you will just find a relationship between the b and c coefficients.

 

[nightweaver066]

hmmmm so you should get these equations?
3a + 4b + 4c= 0
27a + 12b + 4c = 0
4a + b + 2c + 4d - 4 = 0
27a + 18b + 12c + 8d + 8= 0

lol now i dont know which ones to eliminate and stuff :S

Correct except for the third one.

Subbing in (1/2, 1)




So now the equations we have are:





Start off by subtracting (3) from (4). Should flow on from there.

 

[interesting]

For the second part, you would need to use the first derivative:



As the question gave you the turning points, and these points are where the gradients are = 0, simply sub in the x values to obtain 2 equations.

Now you sub in the turning points in to the equation of the curve to obtain another 2 equations.

Using simultaneous equations, the answers will work out nicely to be:


.. if my working out was correct.

how do you know which ones go with which when u simultaneous & did you get these equations?
3a+4b+4c=0
27a+12b+4c=0
a+2b+4c+8d-8=0
27a+18b+12c+8d+8=0

Start off by subtracting (3) from (4). Should flow on from there.

do you have to make them both equal zero before you simultanoeus?

 

[SpiralFlex]

how do you know which ones go with which when u simultaneous & did you get these equations?
3a+4b+4c=0
27a+12b+4c=0
a+2b+4c+8d-8=0
27a+18b+12c+8d+8=0


do you have to make them both equal zero before you simultanoeus?

Since we see all c's in equation 1 and 2, it would be wise to make equation 3 containing three variables a, b, c too. By inspecting the equations, the only way to make all three equations containing these variables is to eliminate d.

 

[interesting]

cool i get it thanks!