If the graph of y = ax^3 + bx^2 + cx + d is s | vedclass

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Question

(C) For a cubic polynomial $y = ax^3 + bx^2 + cx + d$ to be symmetric about a vertical line $x = k$,the coefficient of the cubic term must be zero,i.e

Vedclass · Accepted Answer

$a + \frac{c}{2b} + k = 0$

Vedclass · Answer

(C) For a cubic polynomial $y = ax^3 + bx^2 + cx + d$ to be symmetric about a vertical line $x = k$,the coefficient of the cubic term must be zero,i.e.,$a = 0$.Then the equation becomes $y = bx^2 + cx + d$,which is a parabola.$A$ parabola $y = bx^2 + cx + d$ is symmetric about its axis of symmetry,given by $x = -\frac{c}{2b}$.Given that the graph is symmetric about $x = k$,we have $k = -\frac{c}{2b}$.This implies $2bk = -c$,or $2bk + c = 0$.Since $a = 0$,we can write $a + \frac{c}{2b} + k = 0 + \frac{c}{2b} + (-\frac{c}{2b}) = 0$.Thus,the condition is $a + \frac{c}{2b} + k = 0$.

If the graph of $y = ax^3 + bx^2 + cx + d$ is symmetric about the line $x = k$,then:

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