The equation that represents the 3rd degree polynomial shown on the graph is obtained as follows:
- Because the curve "touches" the x- axis at one point where x = -2, and it "crosses" the x-axis at the point where x = 1, this means that there are two roots at x = -2 and just one root at x = 1.
Therefore, the equation of the polynomial can be generally represented as:
[tex]y=A\times(x+2)^2\times(x-1)[/tex]Now, we need to find the value of the constant A.
To do this simply, we have to look at the y-intercept of the curve. That is, the point where the curve crosses the y-axis. The coordinate of this point is (0, -4), as seen on the graph.
This means that when x = 0, y = -4. We substitute these values into the general equation we obtained earlier, as follows:
[tex]\begin{gathered} y=A\times(x+2)^2\times(x-1) \\ \Rightarrow-4=A\times(0+2)^2\times(0-1) \end{gathered}[/tex]Simplifying the above gives:
[tex]\begin{gathered} -4=A\times(0+2)^2\times(0-1) \\ \Rightarrow-4=A\times(2)^2\times(-1) \\ \Rightarrow-4=A\times4\times-1 \end{gathered}[/tex][tex]\begin{gathered} -4=-4A \\ \Rightarrow\frac{-4}{-4}=A \\ \Rightarrow1=A \\ \Rightarrow A=1 \end{gathered}[/tex]Therefore, we can say that the equation that represents the polynomial shown on the graph is :
[tex]\begin{gathered} y=1\times(x+2)^2\times(x-1) \\ \Rightarrow y=(x+2)^2(x-1) \\ \end{gathered}[/tex]Final simplification gives:
[tex]\begin{gathered} \Rightarrow y=(x+2)^2(x-1) \\ \Rightarrow y=(x^2+4x+4)^{}(x-1) \\ \Rightarrow y=(x^2+4x+4)^{}(x)-(1)(x^2+4x+4)^{} \\ \Rightarrow y=x^3+4x^2+4x-x^2-4x-4 \\ \Rightarrow y=x^3+3x^2-4 \end{gathered}[/tex]Since we are asked to express the polynomial equation in the form of P(x), we can therefore write it as follows:
[tex]P(x)=x^3+3x^2-4[/tex]
