A circle is defined by an equation in the format .

The center is indicated by the point and the radius .

In the equation , the center is and the radius is .

The polynomial is in standard form of a parabola.

To determine the vertex, first write the formula.

Substitute the coefficients.

Since the is negative is negative, the parabola opens down, and we will have a maximum.

The answer is:

A function has to pass the vertical line test, which means that a vertical line can only cross the function one time. To put it another way, for any given value of , there can only be one value of . For the function , there is one value for two possible values. For instance, if , then . But if , as well. This function fails the vertical line test. The other functions listed are a line,, the top half of a right facing parabola, , a cubic equation, , and a semicircle, . These will all pass the vertical line test.

A function has to pass the vertical line test, which means that a vertical line can only cross the function one time. To put it another way, for any given value of , there can only be one value of . For the function , there is one value for two possible values. For instance, if , then . But if , as well. This function fails the vertical line test. The other functions listed are a line,, the top half of a right facing parabola, , a cubic equation, , and a semicircle, . These will all pass the vertical line test.

A circle is defined by an equation in the format .

The center is indicated by the point and the radius .

In the equation , the center is and the radius is .

To find the y-intercept of any line, we must get the equation into the form

where m is the slope and b is the y-intercept.

To manipulate our equation into this form, we must solve for y. First, we must move the x term to the right side of our equation by subtracting it from both sides.

To isolate y, we now must divide each side by 3.

Now that our equation is in the desired form, our y-intercept is simply

We are looking for a point, , where these two lines intersect. While there are many ways to solve for and given two equations, the simplest way I see is to use the elimination method since by adding the two equations together, we can eliminate the variable.

Dividing both sides by 7, we isolate y.

Now, we can plug y back into either equation and solve for x.

Next, we can solve for x.

Therefore, the point where these two lines intersect is .

To calculate a line passing through two points, we first need to calculate the slope, .

Now that we have the slope, we can plug it into our equation for a line in slope intercept form.

To solve for , we can plug in one of the points we were given. For the sake of this example, let's use , but realize either point will give use the same answer.

Now that we have solved for b, we can plug that into our slope intercept form and produce and the answer

The -coordinate of the vertex of a parabola of the form

is

.

Set :

The -coordinate is therefore :

, which is the correct choice.

Set and solve for :

The terms have a GCF of 2, so

The trinomial in parentheses can be FOILed out by noting that and :

And you can divide both sides by 3 to get rid of the coefficient:

Set each of the linear binomials to 0 and solve for :

or

The parabola has as its two intercepts the points and .