91Ó°ÊÓ

The given graph is

An even degree polynomial are either putting up on both ends or down on both ends.

Hence, degree of polynomial is even.

From the graph we can show the beginning and the end of the graph are pointing upward, hence it is opening upwards and thus the coefficient is positive.

An even function is defined when $f\left(x\right)=f(-x)$. Since this function is symmetric about y-axis. Hence, we know the value on the left side of the y-axis equals the value on the right side of y-axis for a given $x$.

Hence, the function is even.

Looking at the function, the graph touches the x-axis 7 times, however in part (a) we conclude that this is an even-degree polynomial.

Since, the graph touches $(0,0)$which means 0 is a zero for this function with at least 2 multiplicty.

Thus,$x^2$must be a factor.

Looking at the function, the graph touches the x-axis 7 times, however in part (a) we conclude that this is an even-degree polynomial.

Hence, minimum degree of polynomial is 8.

To come up with the polynomial, make sure each polynomial contains at a term $x^2$and the roots for one pair of term are opposite as well as three positive and three negative roots.

$$\begin{gathered} y_1=x^2(x-1)(x-2)(x-3)(x+1)(x+2)(x+3) \\ {y}_2=x^2(x-2)(x-3)(x-4)(x+2)(x+3)(x+4) \\ {y}_3=x^2(x-3)(x-4)(x-5)(x+3)(x+4)(x+5) \\ {y}_4=x^2(x-2)(x-4)(x-6)(x+2)(x+4)(x+6) \\ {y}_5=x^2(x-1)(x-3)(x-5)(x+1)(x+3)(x+5) \end{gathered}$$