Question

Which graph best represents the function f(x) = (x + 1)(x − 1)(x − 4)? Graph of a cubic polynomial that falls to the left and rises to the right with x intercepts negative 3, 2, and 3. The graph intersects the y axis at a point between 10 and 15. Graph of a cubic polynomial that falls to the left and rises to the right with x intercepts negative 3, 2, and 3. The graph intersects the y axis at a point between 15 and 20. Graph of a cubic polynomial that falls to the left and rises to the right with x intercepts negative 3, 1, and 3. The graph intersects the y axis at a point between 5 and 10. Graph of a cubic polynomial that falls to the left and rises to the right with x intercepts negative 1, 1, and 4. The graph intersects the y axis at a point between 0 and 5.

Answer 1

`\\ \rm\rightarrowtail y=(x+1)(x-1)(x-4)`

• x intercepts=-1,1,4

Simplify

`\\ \rm\rightarrowtail y=(x^2-1)(x-4)`

`\\ \rm\rightarrowtail y=x^3-4x^2-x+4`

Graph it

behaviour:-

• Left:-Fall
• Right:-rise

Answer 2

Graph of a cubic polynomial that falls to the left and rises to the right with x intercepts negative 1, 1, and 4. The graph intersects the y axis at a point between 0 and 5.

Explanation:

`f(x) = (x + 1)(x- 1)(x-4)`

The x-intercepts are when
`f(x)=0`

`\implies (x+1)=0 \implies x=-1`

`\implies (x-1)=0 \implies x=1`

`\implies (x-4)=0 \implies x=4`

Therefore, the x-intercepts are -1, 1 and 4

The y-intercept is when
`x=0`

`\implies f(0) = (0 + 1)(0- 1)(0-4)=4`

Therefore, the graph intersects the y-axis at (0, 4)

`f(x) = (x + 1)(x- 1)(x-4)`

`\implies f(x)=x^3-4x^2-x+4`

As the degree of the polynomial is odd (3 = cubic) and the leading coefficient is positive, the end behavior of the function is:

`f(x) \rightarrow - \infty, \textsf {as } x \rightarrow - \infty`

`f(x) \rightarrow + \infty, \textsf {as } x \rightarrow + \infty`

So this graph falls to the left and rises to the right.