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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:

Similar Questions

The real root of the equation $x^{3}-6x+9=0$ is

The equation $16x^4 + 16x^3 - 4x - 1 = 0$ has a multiple root. If $\alpha, \beta, \gamma, \delta$ are the roots of this equation,then $\frac{1}{\alpha^4} + \frac{1}{\beta^4} + \frac{1}{\gamma^4} + \frac{1}{\delta^4} =$

Let $f(x)$ be a quadratic polynomial with $f(2)=10$ and $f(-2)=-2$. Then,the coefficient of $x$ in $f(x)$ is

Let $p(x)$ be a quadratic polynomial with real coefficients. If $p(x)=0$ has only purely imaginary roots,then the zeroes of the polynomial $p(p(x))$ are

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4-4x^3+3x^2+2x-2=0$ such that $\alpha$ and $\beta$ are integers and $\gamma, \delta$ are irrational numbers,then $\alpha+2\beta+\gamma^2+\delta^2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module