Answer:
[tex]f(x)=2(x+2)^2(x-1)(x-4)[/tex]
Step-by-step explanation:
The zeros of a function are the x-values of the points at which the curve intersects the x-axis.
From inspection of the given graph, the polynomial has zeros at:
- x = -2 with multiplicity 2 (as the curve touches the x-axis).
- x = 1
- x = 4
The end behaviour of the function is:
[tex]\textsf{As $x \rightarrow -\infty, f(x) \rightarrow +\infty$}[/tex]
[tex]\textsf{As $x \rightarrow +\infty, f(x) \rightarrow +\infty$}[/tex]
This means that:
- The degree of the function is even.
- The leading coefficient is positive.
Therefore:
[tex]\implies f(x)=a(x+2)^2(x-1)(x-4)[/tex]
Substitute the given point (-1, 20) into the function and solve for a:
[tex]\begin{aligned} f(-1)&=20\\a(-1+2)^2(-1-1)(-1-4)&=20\\a(1)(-2)(-5)&=20\\10a&=20\\\implies a&=2\end{aligned}[/tex]
Therefore, the formula for the function is:
[tex]\boxed{f(x)=2(x+2)^2(x-1)(x-4)}[/tex]
In standard form:
[tex]f(x)=2x^4-2x^3-24x^2-8x+32[/tex]
