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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)` **

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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

Judy Johnson

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