Question

The polynomial function graphed here has a leading coefficient (a) that is not 1 or -1. Describe how you would use the information from the graph to write the equation and figure out what the leading coefficient is. Write the entire equation of the polynomial including the leading coefficient.

Answer 1

`\textsf{Factored form:}\quad f(x)=-3(x+1)^2(x-1)`

`\textsf{Standard form:}\quad f(x)=-3x^3-3x^2+3x+3`

Explanation:

The graph shows a cubic function that begins in quadrant II, exhibits a upward concavity, then transitions to a downward concavity, ending in quadrant IV. This indicates that the leading coefficient is negative.

The curve touches the x-axis at x = -1, which implies that there is a root at x = -1. Therefore, the function includes the factor (x + 1). Given that the curve rebounds from the x-axis at this point, this factor will have a multiplicity of 2, making it (x + 1)².

The curve crosses the x-axis at x = 1, which implies there is a root at x = 1. Therefore, the function includes the factor (x - 1). Given that the curve crosses the x-axis at the point, this factor will have a multiplicity of 1, making it (x - 1).

Therefore, the factored form of the equation of the polynomial with leading coefficient "a" is:

`f(x)=-a(x+1)^2(x-1)`

To find the value of "a", substitute the y-intercept (0, 3) into the equation:

`\begin{aligned}f(0)&=3\\-a(0+1)^2(0-1)&=3\\-a(1)^2(-1)&=3\\-a(1)(-1)&=3\\a&=3\end{aligned}`

Therefore, the equation of the graphed polynomial function in factored form is:

`f(x)=-3(x+1)^2(x-1)`

To write it in standard form f(x) = ax³ + bx² + cx + d, expand the brackets:

`f(x)=-3(x+1)(x+1)(x-1)`

`f(x)=-3(x^2+2x+1)(x-1)`

`f(x)=-3(x^3-x^2+2x^2-2x+x-1)`

`f(x)=-3(x^3+x^2-x-1)`

`f(x)=-3x^3-3x^2+3x+3`