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Polynomial Function (Graph/ Inequality)



(CIE 0606/2021/m/22/Q3)
The diagram shows the graph of $y=\mathrm{f}(x)$, where $\mathrm{f}(x)=a(x+b)^{2}(x+c)$ and $a, b$ and $c$ are integers.

(a) Find the value of each of $a, b$ and $c$.

(b) Hence solve the inequality $\mathrm{f}(x) \leqslant-1$ $[3]$

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*********math solution*************

$\begin{array}{ll}&\text { Since } x=-1,-1,1 \text { are zeros of } f(x), \\&y=a(x+1)^{2}(x-1) \cdot \\&(0,-2) \text { lies on the graph, }-2=a(0+1)^{2}(0-1)=-a \\&\therefore a=2 \\&\therefore y=2(x+1)^{2}(x-1) \\&\therefore a=2, b=1, c=-1 \\&\text { From the graph, the value of } x \text { for which } f(x)=-1 \text { is } \\&\qquad x=-1.45,-0.4 \text { and } x=0.85 . \\&\text { Hence the values of } x \text { forwhich } f(x) \leqslant-1 \text { are } \\&x<-1.45 \text { or }-0.4<x<0.85 .\end{array}$
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