Question

Find the formula for the graph attached, given that it is a polynomial, that all zeros of the polynomial are shown, that the exponents of each of the zeros are the least possible, and that it passes through the point (−1, 20)

f(x) = ?

Answer 1

`f(x)=2(x+2)^2(x-1)(x-4)`

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:

`\textsf{As $x \rightarrow -\infty, f(x) \rightarrow +\infty$}`

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

This means that:

• The degree of the function is even.
• The leading coefficient is positive.

Therefore:

`\implies f(x)=a(x+2)^2(x-1)(x-4)`

Substitute the given point (-1, 20) into the function and solve for a:

`\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}`

Therefore, the formula for the function is:

`\boxed{f(x)=2(x+2)^2(x-1)(x-4)}`

In standard form:

`f(x)=2x^4-2x^3-24x^2-8x+32`