in the graph below, the coordinates of the indicated point are (-1, 9/16). construct a polynomial function…

Explanation:

Step1: Identify x - intercepts

The x - intercepts of the graph seem to be at $x=-4$, $x = - 2$ and $x=2$. A polynomial function with these x - intercepts has the form $f(x)=a(x + 4)(x + 2)(x - 2)$. Since the graph touches the x - axis at one of the intercepts, we need a factor with an even power.

Step2: Substitute the given point

We know the point $(-1,\frac{9}{16})$ lies on the graph. Substitute $x=-1$ into the polynomial forms. For $f(x)=a(x + 4)(x + 2)(x - 2)$, we get $f(-1)=a(-1 + 4)(-1 + 2)(-1 - 2)=a(3)(1)(-3)=-9a$. We want $f(-1)=\frac{9}{16}$, so $-9a=\frac{9}{16}$, then $a=-\frac{1}{16}$. Also, since the graph has the general shape (falls to the left and falls to the right), the leading - coefficient must be negative. Let's check the options. Option A: $f(x)=-\frac{1}{16}(x + 4)^2(x + 2)(x - 2)$. Substitute $x=-1$: [ \begin{align*} f(-1)&=-\frac{1}{16}(-1 + 4)^2(-1 + 2)(-1 - 2)\ &=-\frac{1}{16}(9)(1)(-3)\ &=\frac{27}{16} \end{align*} ] Option B: $f(x)=\frac{1}{16}(x + 4)(x + 2)(x - 2)^2$. The leading - coefficient is positive, so it can't be this one as the graph falls to the left and right. Option C: $f(x)=\frac{1}{16}(x + 4)(x + 2)^2(x - 2)$. The leading - coefficient is positive, so it's not this one. Option D: $f(x)=-\frac{1}{16}(x + 4)(x + 2)(x - 2)$. Substitute $x=-1$: [ \begin{align*} f(-1)&=-\frac{1}{16}(-1 + 4)(-1 + 2)(-1 - 2)\ &=-\frac{1}{16}(3)(1)(-3)\ &=\frac{9}{16} \end{align*} ] Option E: $f(x)=-\frac{1}{16}(x + 4)(x + 2)^2(x - 2)$. Substitute $x=-1$: [ \begin{align*} f(-1)&=-\frac{1}{16}(-1 + 4)(-1 + 2)^2(-1 - 2)\ &=-\frac{1}{16}(3)(1)(-3)\ &=\frac{9}{16} \end{align*} ] Option F: $f(x)=-\frac{1}{16}(x + 4)(x + 2)(x - 2)^2$. Substitute $x=-1$: [ \begin{align*} f(-1)&=-\frac{1}{16}(-1 + 4)(-1 + 2)(-1 - 2)^2\ &=-\frac{1}{16}(3)(1)(9)\ &=-\frac{27}{16} \end{align*} ] We want the smallest - degree polynomial. The polynomial $f(x)=-\frac{1}{16}(x + 4)(x + 2)(x - 2)$ is of degree 3, while the polynomials with squared factors are of degree 4.

Answer:

D. $f(x)=-\frac{1}{16}(x + 4)(x + 2)(x - 2)$