Expert in study
alarm
Ask
Other

if the graph of ax^3+bx^2+cx+d=0 is symmetrical about the line x=k ,then what is the value of a+k?? |

Register to add an answer
answers: 1
466 cents
Answer:

If the graph of `f(x)=ax^3+bx^2+cx+d` is symmetrical about the the line `x=k` , then the underlying function is odd with perhaps a horizontal and/or vertical translation. (The symmetry must be rotational)

Thus the general form for such a function is `f(x)=a(x-k)^3+cx+d`

Expanding we get `f(x)=a[x^3-3kx^2+3k^2x-k^3]+cx+d`

`=ax^3-3akx^2+(3ak^2+c)x+(d-ak^3)`

Thus by construction we have `f(x)=ax^3-3akx^2+(3ak^2+c)x+(d-ak^3)` symmetric about the line `x=k` .

*From calculus we can find that the line of symmetry intersects the point of inflection. Considering our original equation, we take the second derivative and set equal to zero to find the point of inflection:

`f(x)=ax^3+bx^2+cx+d`

`f'(x)=3ax^2+2bx+c`

`f''(x)=6ax+2b`

`6ax+2b=0 => x=(-2b)/(6a)=(-b)/(3a)`

Thus the line of symmetry will be at `x=(-b)/(3a)` . But `x=k` is our line of symmetry so `(-b)/(3a)=k => b=-3ak` as above.*

So `a+k=a+(-b)/(3a)=(3a^2-b)/(3a)`

__________________________________________

Example: Let `f(x)=3x^3-9x^2+11x+2` .(`3(x-1)^3+2x+5`) The graph of this function is symmetric about the line `x=1` , so `a+k=3+1=4` and `(3(3)^2-(-9))/(3(3))=36/9=4`

102
For answers need to register.
Contacts
mail@expertinstudy.com
Feedback
Expert in study
About us
For new users
For new experts
Terms and Conditions