First, given the two points, find the equation of the line using the slope formula and the y-intercept equation.

Slope:

\(\displaystyle \frac{y_2-y_1}{x_2-x_1} = \frac{3-(-2)}{1-(-3)}=\frac{5}{4}\)

Write the slope-intercept formula.

\(\displaystyle y=mx+b\)

Substitute a given point and the slope into the equation to find the y-intercept.

\(\displaystyle 3=\left(\frac{5}{4}\right)(1)+b\)

\(\displaystyle \frac{12}{4}=\frac{5}{4}+b\)

\(\displaystyle b=\frac{7}{4}\)

The y-intercept is: \(\displaystyle b=\frac{7}{4}\).

Substitiute the slope and the y-intercept into the slope-intercept form.

\(\displaystyle y=\frac{5}{4}x+\frac{7}{4}\)

To find the x-intercept, substitute \(\displaystyle y=0\) and solve for x.

\(\displaystyle 0=\frac{5}{4}x+\frac{7}{4}\)

\(\displaystyle -\frac{7}{4}=\frac{5}{4}x\)

\(\displaystyle -\frac{7}{4} \cdot (\frac{4}{5})=\frac{5}{4}x \cdot (\frac{4}{5})\)

\(\displaystyle x=-\frac{7}{5}\)

The x-intercept is: \(\displaystyle x=-\frac{7}{5}\)

Write the standard form of the parabola. 

\(\displaystyle y=ax^2+bx+c\)

Given the point \(\displaystyle (0,-4)\), the y-intercept is -4, which indicates that \(\displaystyle c=-4\).  This is also the vertex, so the vertex formula can allow writing an expression in terms of variables \(\displaystyle a\) and \(\displaystyle b\).   

Write the vertex formula and substitute the known vertex given point  \(\displaystyle (0,-4)\).

\(\displaystyle x=-\frac{b}{2a}\)

\(\displaystyle 0=-\frac{b}{2a}\)

\(\displaystyle b=0\)

Using the values of \(\displaystyle b\), \(\displaystyle c\), and the other given point \(\displaystyle (2,0)\), substitute these values to the standard form and solve for \(\displaystyle a\).

\(\displaystyle 0=a(2)^2+(0)(2)+(-4)\)

\(\displaystyle 0=4a-4\)

\(\displaystyle 4=4a\)

\(\displaystyle a=1\)

Substitute the values of \(\displaystyle a\),\(\displaystyle b\), and \(\displaystyle c\) into the standard form of the parabola.

The correct answer is:  \(\displaystyle y=x^2-4\)

Write the slope intercept formula.

\(\displaystyle y=mx+b\)

Convert the given x-intercept to a known point, which is \(\displaystyle (1,0)\).

Substitute the given slope and the point to solve for the y-intercept.

\(\displaystyle 0=(1)(1)+b\)

\(\displaystyle b=-1\)

Substitute the slope and y-intercept into the slope-intercept formula.

\(\displaystyle y=x-1\)

Add 1 on both sides of the equation, and subtract \(\displaystyle y\) on both sides of the equation to find the equation in standard form.

\(\displaystyle x-y=1\)

By the Fundamental Theorem of Algebra, a polynomial equation of degree 3 must have three solutions, or roots, but one root can be a double root or triple root. Since the polynomial here has two roots, \(\displaystyle -4\) and 4, one of these must be a double root. Since the leading term is \(\displaystyle x^{2}\), the equation must be 

\(\displaystyle (x- (-4))(x- (-4))(x-4) = 0\)

or 

\(\displaystyle (x- (-4))(x- 4)(x-4) = 0\)

We rewrite both.

\(\displaystyle (x- (-4))(x- (-4))(x-4) = 0\)

\(\displaystyle (x+4)(x+4)(x-4) = 0\)

\(\displaystyle (x+4)(x^{2}-4^{2}) = 0\)

\(\displaystyle (x+4)(x^{2}-16) = 0\)

\(\displaystyle x \cdot x^{2}- x \cdot 16 + 4 \cdot x^{2}- 4 \cdot 16 = 0\)

\(\displaystyle x^{3}-16 x + 4 x^{2}- 64 = 0\)

\(\displaystyle x^{3}+ 4 x^{2}-16 x - 64 = 0\)

\(\displaystyle (x- (-4))(x- 4)(x-4) = 0\)

\(\displaystyle (x+4)(x- 4)(x-4) = 0\)

\(\displaystyle (x^{2}-16) (x-4)= 0\)

\(\displaystyle x \cdot x^{2}- 4 \cdot x^{2}- x \cdot 16 + 4 \cdot 16 = 0\)

\(\displaystyle x^{3}- 4 x^{2}-16 x + 64 = 0\)

The correct response can be  \(\displaystyle f(x) = x^{3}+ 4 x^{2}-16 x - 64\) or \(\displaystyle f(x) = x^{3}- 4 x^{2}-16 x + 64\). The first is not among the choices, so the last is the correct choice.

We can evaluate \(\displaystyle f(-5)\) in each of the definitions of \(\displaystyle f\) in the five choices. If \(\displaystyle f(-5) = 0\), \(\displaystyle (-5,0)\) is an \(\displaystyle x\)-intercept.

\(\displaystyle f(x) = x^{3}+ 5x^{2} -25 x-125\)

\(\displaystyle f(-5) = (-5)^{3}+ 5 \cdot (-5)^{2} -25 (-5)-125\)

\(\displaystyle = (-5)^{3}+ 5 \cdot 25 - (-125)-125\)

\(\displaystyle = -125 +125 +125-125 = 0\)

\(\displaystyle f(x) = x^{3}- 5x^{2} -25 x+125\)

\(\displaystyle f(-5) = (-5)^{3}- 5 \cdot (-5)^{2} -25 (-5)+125\)

\(\displaystyle = (-5)^{3}+ 5 \cdot 25 - (-125)-125\)

\(\displaystyle = -125 +125 +125-125 = 0\)

\(\displaystyle f(x) = x^{3} +125\)

\(\displaystyle f(-5) = (-5) ^{3} +125\)

\(\displaystyle = -125+125 = 0\)

\(\displaystyle f(x) = x^{3}+1 5x^{2} + 75 x+125\)

\(\displaystyle f(-5) = (-5)^{3}+1 5\cdot (-5)^{2} + 75 (-5)+125\)

\(\displaystyle f(-5) = -125+1 5\cdot 25 + (-375)+125\)

\(\displaystyle f(-5) = -125+375-375+125 = 0\)

\(\displaystyle f(x) = x^{3}-1 5x^{2} + 75 x-125\)

\(\displaystyle f(-5) = (-5)^{3}-1 5\cdot (-5)^{2} + 75 (-5)-125\)

\(\displaystyle f(-5) = -125-1 5\cdot 25 + (-375)-125\)

\(\displaystyle f(-5) = -125-375-375-125 = -1,000\)

\(\displaystyle f(x) = x^{3}-1 5x^{2} + 75 x-125\) does not have \(\displaystyle (-5,0)\) as an \(\displaystyle x\)-intercept, so it is the correct choice.

Since the graph of a function \(\displaystyle f\) has its \(\displaystyle x\)-intercept at a point \(\displaystyle (a,0)\) if and only if \(\displaystyle f(a)= 0\), finding possible \(\displaystyle x\)-intercepts of the graph of \(\displaystyle f\) is equivalent to finding a solution of \(\displaystyle f(x)= 0\). Since \(\displaystyle f\) has integer coefficients, then by the Rational Zeroes Theorem, any rational solutions to the equation

\(\displaystyle f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12\)

must be the quotient, or the (negative) opposite of the quotient, of a factor of constant coefficient 12 - that is, an element of \(\displaystyle \left \{ 1, 2, 3, 4, 6, 12 \right \}\) - and a factor of leading coefficient 2 - that is, an element of \(\displaystyle \left \{ 1,2 \right \}\). Since all of the choices are positive, we will only look at possible positive solutions.

The quotients of an element of the first set and an element of the last are:

\(\displaystyle \frac{1}{1} = 1\); \(\displaystyle \frac{2}{1} = 2\); \(\displaystyle \frac{3}{1} = 3\); \(\displaystyle \frac{4}{1} = 4\); \(\displaystyle \frac{6}{1} = 6\); \(\displaystyle \frac{12}{1} = 12\);

\(\displaystyle \frac{1}{2}\); \(\displaystyle \frac{2}{2} = 1\); \(\displaystyle \frac{3}{2}\); \(\displaystyle \frac{4}{2} = 2\); \(\displaystyle \frac{6}{2} = 3\); \(\displaystyle \frac{12}{2} = 6\)

Eliminating duplicates, the set of possible positive rational solutions to \(\displaystyle f(x) = 0\) is

\(\displaystyle \left \{ \frac{1}{2},1, \frac{3}{2} , 2,3,4,6,12 \right \}\).

Of the five choices, only \(\displaystyle \frac{1}{6}\) does not appear in the set of possible rational solutions of \(\displaystyle f(x) = 0\), so of the five choices, only \(\displaystyle \left ( \frac{1}{6}, 0 \right )\) cannot be an \(\displaystyle x\)-intercept of the graph.

As a polynomial function, \(\displaystyle f\) has a continuous graph. By the Intermediate Value Theorem, if \(\displaystyle f(a)\) and \(\displaystyle f(b)\) are of different sign, then \(\displaystyle f(c) = 0\) for some \(\displaystyle c \in (a,b)\) - that is, the graph of \(\displaystyle f\) has an \(\displaystyle x\)-intercept between \(\displaystyle (a,0)\) and \(\displaystyle (b,0)\). Evaluate \(\displaystyle f\) for all \(\displaystyle x= 0,1,2,3,4,5\) and observe between which two integers the sign changes. 

\(\displaystyle f(x) = x^{4}+3x^{2}-17x-115\)

\(\displaystyle f(0) = 0^{4}+3 \cdot 0^{2}-17\cdot 0-115\)

\(\displaystyle = 0+0-0-115\)

\(\displaystyle = -115\)

\(\displaystyle f(1) =1^{4}+3\cdot 1^{2}-17 \cdot 1-115\)

\(\displaystyle = 1 +3 -17-115\)

\(\displaystyle = -128\)

\(\displaystyle f(2) = 2^{4}+3\cdot 2^{2}-17\cdot 2-115\)

\(\displaystyle = 16+3\cdot 4 -34 -115\)

\(\displaystyle = 16+12 -34 -115\)

\(\displaystyle = -121\)

\(\displaystyle f(3) =3 ^{4}+3 \cdot 3^{2}-17 \cdot 3-115\)

\(\displaystyle = 81 + 3 \cdot 9 - 51 -115\)

\(\displaystyle = 81 + 27 -51 -115\)

\(\displaystyle = -58\)

\(\displaystyle f(4) = 4 ^{4}+3\cdot 4^{2}-17\cdot 4 -115\)

\(\displaystyle = 256 +3\cdot 16 -68-115\)

\(\displaystyle = 256 +48 -68-115\)

\(\displaystyle 121\)

\(\displaystyle f(5) = 5^{4}+3 \cdot 5^{2}-17 \cdot 5 -115\)

\(\displaystyle = 625 +3 \cdot 25 -85 -115\)

\(\displaystyle = 625 +75 -85 -115\)

\(\displaystyle =500\)

Since \(\displaystyle f(3)< 0\) and \(\displaystyle f(4) > 0\), the \(\displaystyle x\)-intercept is between \(\displaystyle (3,0)\) and \(\displaystyle (4,0)\).

The Intermediate Value Theorem states that if \(\displaystyle f\) is a continuous function, as all five of the polynomial functions in the given choices are, and \(\displaystyle f\left ( \frac{1}{2} \right )\) and \(\displaystyle f(1)\) are of different sign, then the graph of \(\displaystyle f\) has an \(\displaystyle x\)-intercept on the interval \(\displaystyle \left ( \frac{1}{2}, 1 \right )\).

We evaluate \(\displaystyle f\left ( \frac{1}{2} \right )\) and \(\displaystyle f(1)\) for each of the five choices to find the one for which the two have different sign.

\(\displaystyle f(x) = x^{4}-x^{3}+ x-3\)

\(\displaystyle f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+ \frac{1}{2} -3 = \frac{1}{16} - \frac{1}{8}+ \frac{1}{2}-3=-2\frac{9}{16}\)

\(\displaystyle f(1) = 1^{4}-1^{3}+ 1-3 = 1-1+1-3 = -2\)

\(\displaystyle f\left ( \frac{1}{2} \right )\) and \(\displaystyle f(1)\) are both negative.

\(\displaystyle f(x) = x^{4}-x^{3}+ 2 x-3\)

\(\displaystyle f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+2 \left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+1-3=-2\frac{1}{16}\)

\(\displaystyle f(1) = 1^{4}-1^{3}+ 2 \cdot 1-3 = 1-1+2-3 = -1\)

\(\displaystyle f\left ( \frac{1}{2} \right )\) and \(\displaystyle f(1)\) are both negative.

\(\displaystyle f(x) = x^{4}-x^{3}+ 4 x-3\)

\(\displaystyle f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+4 \left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+2-3=-1\frac{1}{16}\)

\(\displaystyle f(1) = 1^{4}-1^{3}+ 4 \cdot 1-3 = 1-1+4-3 = 1\)

\(\displaystyle f\left ( \frac{1}{2} \right )\) and \(\displaystyle f(1)\) are of different sign.

\(\displaystyle f(x) = x^{4}-x^{3}+ 8 x-3\)

\(\displaystyle f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+8\left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+4-3= \frac{15}{16}\)

\(\displaystyle f(1) = 1^{4}-1^{3}+ 8 \cdot 1-3 = 1-1+8-3 = 5\)

\(\displaystyle f\left ( \frac{1}{2} \right )\) and \(\displaystyle f(1)\) are both positive.

\(\displaystyle f(x) = x^{4}-x^{3}+ 16x-3\)

\(\displaystyle f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+16\left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+8-3=4 \frac{15}{16}\)

\(\displaystyle f(1) = 1^{4}-1^{3}+ 16 \cdot 1-3 = 1-1+16-3 = 13\)

\(\displaystyle f\left ( \frac{1}{2} \right )\) and \(\displaystyle f(1)\) are both positive.

 \(\displaystyle f(x) = x^{4}-x^{3}+ 4 x-3\) is the function in which \(\displaystyle f\left ( \frac{1}{2} \right )\) and \(\displaystyle f(1)\) are of different sign, so it is represented by a graph with an \(\displaystyle x\)-intercept between \(\displaystyle \left ( \frac{1}{2},0 \right )\) and \(\displaystyle (1 ,0)\). This is the correct choice.

A polynomial equation of degree 3 with solution set \(\displaystyle \left \{ -2, 2, 3\right \}\) and leading term \(\displaystyle x^{2}\) takes the form

\(\displaystyle (x- (-2)) (x-2)(x-3)= 0\)

We can rewrite this as follows:

\(\displaystyle (x+2) (x-2)(x-3)= 0\)

\(\displaystyle (x^{2}-2^{2})(x-3)= 0\)

\(\displaystyle (x^{2}-4)(x-3)= 0\)

\(\displaystyle x^{2}\cdot x- x^{2} \cdot 3 - 4\cdot x + 4 \cdot 3 = 0\)

\(\displaystyle x^{3} - 3 x^{2} - 4 x + 12 = 0\)

The correct response is \(\displaystyle f(x)= x^{3} - 3 x^{2} - 4 x + 12\).