It is not necessary to actually find the solutions to a quadratic equation to determine the sum of its solutions.

First, get the equation in standard form \(\displaystyle ax^{2}+ bx + c = 0\) by subtracting \(\displaystyle 17x\) from both sides:

\(\displaystyle x^{2} + 34 = 17x\)

\(\displaystyle x^{2} + 34 -17x = 17x - 17x\)

\(\displaystyle x^{2} -17x + 34 = 0\)

If a quadratic equation has two distinct solutions, which we are given here, their sum is the linear coefficient \(\displaystyle b\). In this problem, \(\displaystyle b= -17\), making \(\displaystyle -17\) the correct choice.

The Intermediate Value Theorem (IVT) states that if the graph of a function \(\displaystyle f\) is continuous on an interval \(\displaystyle [a, b]\), and \(\displaystyle f(a)\) and \(\displaystyle f(b)\) differ in sign, then \(\displaystyle f\) has a zero on \(\displaystyle (a,b)\). Consequently, the way to answer this question is to determine the signs of \(\displaystyle f\) on the endpoints of the subintervals - \(\displaystyle \left \{ 0, 0.2, 0.4, 0.6, 0.8, 1 \right \}\). We can do this by substituting each value for \(\displaystyle x\) as follows:

\(\displaystyle f(x)= x^{3}+x^{2}-5x+2\)

\(\displaystyle f(0)= 0^{3}+0^{2}-5(0)+2 = 0+0-0+2= 2\)

\(\displaystyle f(0.2)= 0.2^{3}+0.2^{2}-5(0.2)+2 = 0.008+0.04-1+2 = 1.048\)

\(\displaystyle f(0.4)= 0.4^{3}+0.4^{2}-5(0.4)+2= 0.064+ 0.16- 2+2 = 0.224\)

\(\displaystyle f(0.6)= 0.6^{3}+0.6^{2}-5(0.6)+2= 0.216+ 0.36- 3+2 = -0.424\)

\(\displaystyle f(0.8)= 0.8^{3}+0.8^{2}-5(0.8)+2= 0.512+ 0.64- 4+2 = -0.848\)

\(\displaystyle f(1)=1^{3}+1^{2}-5(1)+2 =1+1-5+2=-1\)

\(\displaystyle f\) assumes positive values for \(\displaystyle x= 0, 0.2, 0.4\) and negative values for \(\displaystyle x = 0.6, 0.8, 1\). By the IVT, \(\displaystyle f\) has a zero on \(\displaystyle (0.4, 0.6)\).

Use the zeroes of the graph to determine the matching function. Zeroes are values of x where \(\displaystyle f(x)=0\). In other words, they are points on the graph where the curve touches zero. 

Visually, you can see that the curve crosses the x-axis when \(\displaystyle x=-2\), \(\displaystyle x=-1\), and \(\displaystyle x=2\). Therefore, you need to look for a function that will equal zero at these x values. 

A function with a factor of \(\displaystyle (x+2)\) will equal zero when  \(\displaystyle x=-2\), because the factor of \(\displaystyle (x+2)\) will equal zero. The matching factors for the other two zeroes, \(\displaystyle x=-1\) and \(\displaystyle x=2\), are \(\displaystyle (x+1)\) and \(\displaystyle (x-2)\), respectively.  

The answer choice \(\displaystyle f(x)=-2x^2(x+1)(x-2)(x+2)\) has all of these factors, but it is not the answer because it has an additional zero that would be visible on the graph. Notice it has a factor of \(\displaystyle x^2\), which results in a zero at \(\displaystyle x=0\). This additional zero that isn't present in the graph indicates that this cannot be matching function.  

\(\displaystyle f(x)=-2(x+1)(x-2)(x+2)\) is the answer because it has all of the required factors and, as a result, the required zeroes, while not having additional zeroes. Notice that the constant coefficient of negative 2 does not affect where the zeroes are. 

First, look at the zeroes of the graph. Zeroes are where the function touches the x-axis (i.e. values of \(\displaystyle x\) where \(\displaystyle f(x)=0\)). 

The graph shows the function touching the x-axis when \(\displaystyle x=-2\), \(\displaystyle x=0\), and at a value in between 1.5 and 2.

Notice all of the possible answers are already factored. Therefore, look for one with a factor of \(\displaystyle x\) (which will make \(\displaystyle f(x)=0\) when \(\displaystyle x=0\)), a factor of \(\displaystyle (x+2)\) to make \(\displaystyle f(x)=0\) when \(\displaystyle x=-2\), and a factor which will make \(\displaystyle f(x)=0\) when \(\displaystyle x\) is at a value between 1.5 and 2.

\(\displaystyle f(x)= x(3x-5)(x+2)\)

This function fills the criteria; it has an \(\displaystyle x\) and an \(\displaystyle (x+2)\) factor. Additionally, the third factor, \(\displaystyle (3x-5)\), will result in \(\displaystyle f(x)=0\) when \(\displaystyle x=\frac{5}{3}\), which fits the image. It also does not have any extra zeroes that would contradict the graph.