Write the standard form of the parabola. 

Given the point , the y-intercept is -4, which indicates that .  This is also the vertex, so the vertex formula can allow writing an expression in terms of variables  and .   

Write the vertex formula and substitute the known vertex given point  .

Using the values of , , and the other given point , substitute these values to the standard form and solve for .

Substitute the values of ,, and  into the standard form of the parabola.

The correct answer is:  

Write the slope intercept formula.

Convert the given x-intercept to a known point, which is .

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

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

Add 1 on both sides of the equation, and subtract  on both sides of the equation to find the equation in standard form.

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,  and 4, one of these must be a double root. Since the leading term is , the equation must be 

or 

We rewrite both.

The correct response can be   or . The first is not among the choices, so the last is the correct choice.

We can evaluate  in each of the definitions of  in the five choices. If ,  is an -intercept.

 does not have  as an -intercept, so it is the correct choice.

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

must be the quotient, or the (negative) opposite of the quotient, of a factor of constant coefficient 12 - that is, an element of  - and a factor of leading coefficient 2 - that is, an element of . 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:

; ; ; ; ; ;

; ; ; ; ;

Eliminating duplicates, the set of possible positive rational solutions to is

.

Of the five choices, only does not appear in the set of possible rational solutions of , so of the five choices, only cannot be an -intercept of the graph.

As a polynomial function,  has a continuous graph. By the Intermediate Value Theorem, if  and  are of different sign, then  for some  - that is, the graph of  has an -intercept between  and . Evaluate  for all  and observe between which two integers the sign changes. 

Since  and , the -intercept is between  and .

The Intermediate Value Theorem states that if  is a continuous function, as all five of the polynomial functions in the given choices are, and  and  are of different sign, then the graph of  has an -intercept on the interval .

We evaluate  and  for each of the five choices to find the one for which the two have different sign.

 and  are both negative.

 and  are both negative.

 and  are of different sign.

 and  are both positive.

 and  are both positive.

  is the function in which  and  are of different sign, so it is represented by a graph with an -intercept between  and . This is the correct choice.

A polynomial equation of degree 3 with solution set  and leading term  takes the form

We can rewrite this as follows:

The correct response is .

The equation of a circle with center  and radius  is 

The center is the midpoint of any diameter, so to find the center, we use the midpoint formula:

The center is (7,3). The radius is the distance between (7,3) and (10,1), so we use the distance formula:

So , and the equation of the circle is

The distance of the two points is .

So 10 is the radius of the circle. Then we can calculate the area: 

.