The function h is given by $h(x) = ax^3 + bx^2 +cx +d$.

where a,b,c and d are real constants

The graph of $y =h(x)$ passes through the points $(1,1)$ and $(2,2)$. Given that $(2,2)$ is a maximum point , find three linear equations involving a,b,c and d.

It is given further that the point (-4 ,14) lies on the graph of y=|h(x)| , and that h(x) is strictly decreasing for $x \geq 2$ . Find the values of a,b,c and d.

I tried to find 3 required equations , but the results,which I found , are not satisfied with the requirement . I hope that I would be given a nice method to solve these questions . Thank you !

  • $\begingroup$ May I suggest you show what you already did ? $\endgroup$ Sep 16, 2017 at 13:23
  • $\begingroup$ I just replace y and x as the coordinate of 2 points which h(x) passes .Those are all I can do $\endgroup$ Sep 16, 2017 at 14:00

2 Answers 2


Considering $$h(x) = ax^3 + bx^2 +cx +d$$ apply the formula to get $$a+b+c+d=1\tag 1$$ $$8a+4b+2c+d=1\tag 2$$ You are also told that the function is maximum at $(2,2)$ which means that $$h'(x)=3ax^2+2bx+c$$ is zero at this point. So, $$12a+4b+c=0\tag 3$$ Is it so complex ?


You have $a+b+c+d=1$ and $8a+4b+2c+d=2$.

Because $(2,2)$ is maximum point, so $h'(2) = 12a + 4b + c = 0$.

Point (-4,14) lies on $|h(x)|$, then $14 = |-64a+16b-4c+d|$.

From here, you have two cases.

Case 1 $\begin{cases}a+b+c+d&=&1\\8a+4b+2c+d&=&2\\12a+4b+c&=&0\\-64a+16b-4c+d&=&14\end{cases}$

Case 1 $\begin{cases}a+b+c+d&=&1\\8a+4b+2c+d&=&2\\12a+4b+c&=&0\\-64a+16b-4c+d&=&-14\end{cases}$

Solve it, you got $a,b,c,d$.

Because $h(x)$ decreases strictly for $x\geq 2$, so $h'(x) < 0$ for all $x\geq 2$, or $$3ax^2 + 2bx + c < 0.$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .