Does y = ax^3 + bx^2 + cx + d Have Two Distinct Turning Points if b^2 > 3ac?

In summary, the graph of y= ax^3 + bx^2 + cx + d has two distinct turning points if b^2 > 3ac. To find values of a, b, c, and d for which the graph has turning points at (0.5, 1) and (1.5, -1), the equation is differentiated with respect to x and the maxima and minima are found. The discriminant must be greater than 0 for two distinct turning points. For the second part, the equation 3ax^2 + 2bx + c = 0 is used and the values of a, b, c, and d are solved for using simultaneous equations. The
  • #1
disfused_3289
13
0
Prove that the graph of y= ax^3 + bx^2 + cx + d has two distinct turning points if
b^2> 3ac. Find values of a,b,c and d for which the graph of this form has turning points at (0.5, 1) and (1.5, -1)
 
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  • #2
hi iam a new member to this site

well here isolved first part of it
differentiate the whole equation with respeact to x

you will get

3ax^2+2bx+c

find the maxima and minima

i.e for two turning points two distinct max,min must exist

Discriminant>0

(2b)^2-12ac>0
4*b^2 > 12 ac
b^2 >3 ac...prooved
 
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  • #3
second part

3ax^2+2bx+c=0 was the equation i already mentioned
one answer was 1/2,second 3/2
adding them 2
multiplying them we get 3/4

2=-2b/3a
-b=3a

3/4=c/3a
4c=9a

substituting value of function in the function equation we get

a+2b+4c+8d=8
27a+18b+12c+8d=-8

solve these equations simultaniously
for a,b,c,d

we get

d=-1
a=4
b=-12
c=9

if iam correct
 
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  • #5
I don't understand your last statement ..
please pardon...:smile:
 
  • #6
Solved

Ah, I now understand it. I never thought of delta before
Thanks alot
 

FAQ: Does y = ax^3 + bx^2 + cx + d Have Two Distinct Turning Points if b^2 > 3ac?

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