Question

A cubic polynomial $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ has a graph which is tangent to the x - axis at 2 has another x-intercept at -1 and has y-intercept at -2 as shown The values of a+b+c+d equals

• A. -2
• B. -1
• C. 0
• D. 1

Answer 1

Answer: (B)

-1

Given, $f\left(x\right)=a{x}^3+b{x}^2+cx+d$
$f^{\prime }\left(x\right)=3a{x}^2+2bx+c$
It can be observed from the the given graph,
$f\left(2\right)=0\Rightarrow 8a+4b+2c+d=0...\left(1\right)$
$f(-1)=0\Rightarrow -a+b-c+d=0.......\left(2\right)$
$f\left(0\right)=-2.....\left(3\right)$
and $f^{\prime }\left(2\right)=0\Rightarrow 12a+4b+c=\mathrm{0....}\left(4\right)$
Solving (1),(2),(3)&(4)
We get $a=-\frac{1}{2},b=\frac{3}{2},c=0,d=-2$
Hence, $a+b+c+d=-1$