Question

If a tutor knows how to help me do this please help!!

Answer 1

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:

`y=A*(x+2)^2*(x-1)`

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:

`\begin{gathered} y=A*(x+2)^2*(x-1) \\ \Rightarrow-4=A*(0+2)^2*(0-1) \end{gathered}`

Simplifying the above gives:

`\begin{gathered} -4=A*(0+2)^2*(0-1) \\ \Rightarrow-4=A*(2)^2*(-1) \\ \Rightarrow-4=A*4*-1 \end{gathered}`
`\begin{gathered} -4=-4A \\ \Rightarrow(-4)/(-4)=A \\ \Rightarrow1=A \\ \Rightarrow A=1 \end{gathered}`

Therefore, we can say that the equation that represents the polynomial shown on the graph is :

`\begin{gathered} y=1*(x+2)^2*(x-1) \\ \Rightarrow y=(x+2)^2(x-1) \\ \end{gathered}`

Final simplification gives:

`\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}`

Since we are asked to express the polynomial equation in the form of P(x), we can therefore write it as follows:

`P(x)=x^3+3x^2-4`