Starting with

moves the parabola by units to the right.

Similarly moves the parabola by units to the left.

Hence the correct answer is option .

The highest exponent of the variable term is two (). This tells that this function is quadratic, meaning that it is a parabola.

The graph below will be the answer, as it shows a parabolic curve.

To find where the graph crosses the horizontal axis, we need to set the function equal to 0, since the value at any point along the axis is always zero.

To find the possible rational zeroes of a polynomial, use the rational zeroes theorem:

Our constant is 10, and our leading coefficient is 1. So here are our possible roots:

Let's try all of them and see if they work! We're going to substitute each value in for using synthetic substitution. We'll try -1 first.

Looks like that worked! We got 0 as our final answer after synthetic substitution. What's left in the bottom row helps us factor down a little farther:

We keep doing this process until is completely factored:

Thus, crosses the axis at .

The graph of a quadratic function has an -intercept at any point at which , so we set the quadratic expression equal to 0:

Since the question simply asks for the number of -intercepts, it suffices to find the discriminant of the equation and to use it to determine this number. The discriminant of the quadratic equation

is

.

Set , and evaluate:

The discriminant is positive, so the has two real zeroes - and its graph has two -intercepts.

When we look at the function we see that the highest power of the function is a 3 which means it is an "odd degree" function. This means that the right and left side of the function will approach opposite directions. *Remember O for Odd and O for opposite.

In this case we also have a negative sign associated with the highest power portion of the function - this means that the function is flipped.

Both of these combine to make this an "odd negative" function.

Odd negative functions always have the right side of the function approaching down and the left side approaching up.

We represent this mathematically by saying that as x approaches negative infinity (left side), the function will approach positive infinity:

...and as x approaches positive infinity (right side) the function will approach negative infinity:

As a polynomial function, the graph of is continuous. By the Intermediate Value Theorem (IVT), if or , then there must exist a value such that .

Setting , and examining the first condition, the above becomes:

if , then there must exist a value such that - or, restated, must have a zero on the interval . Since , . the condition holds, and by the IVT, it follows that has a zero on .

The parabola of an equation of the form is vertical, and faces upward or downward depending entirely on the sign of , the coefficient of . This coefficient, , is negative; the parabola is concave downward.

The graph of a quadratic function has an -intercept at any point at which , so, first, set the quadratic expression equal to 0:

The number of -intercepts of the graph is equal to the number of real zeroes of the above equation, which can be determined by evaluating the discriminant of the equation, . Set , and evaluate:

The discriminant is equal to zero, so the quadratic equation has one real zero, and the graph of has exactly one -intercept.

As a polynomial function, the graph of is continuous. By the Intermediate Value Theorem, if or , then there must exist a value such that .

Setting , and , this becomes: If or , then there must exist a value such that - that is, must have a zero on .

However, the question is asking us to use the converse of this statement, which is not true in general. If has a zero on , it does not necessarily follow that or - specifically, with , it does not necessarily follow that . A counterexample is the function shown below, which fits the conditions of the problem but does not have a negative value for :

The answer is false.

The graph of the quadratic function is a parabola with its vertex at the point with coordinates

.

Set ; the -coordinate is

Evaluate by substitution:

The vertex has 0 as its -coordinate; it is therefore on an axis.