We use the distance formula to determine the length of the radius:

Which of the following would give a graph of an upward facing parabola?

Parabolas are created from polynomials where the highest exponent is 2. 

To be an upward facing parabola, the squared term must be positive.

The only option that matches these criteria is:

To find the vertex of this parabola use the following formula to find the x-coordinate of its vertex; find the y-coordinate by substituting the x-coordinate into the equation. 

To find the y-coordinate, substitute -2 back into the quadratic equation:

The vertex is .

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 positive, so the equation has two solutions, both of which are real. Consequently, the graph of the function  has two -intercepts

The highest point of the ball is the vertex of the ball's parabolic path, so to find the number of seconds  that is takes to reach this point, it is necessary to find the first coordinate  of the vertex of the parabola of the graph of the function

The parabola of the graph of 

has as its ordinate, or -coordinate,

,

so, setting ,

,

This is the time in seconds that it takes the ball to reach the highest point of its path.

The parabola of the equation  has its vertex at a point with -coordinate ; set  and  in this formula to get

Substitute this for  in the equation to obtain the corresponding -coordinate:

The vertex is at .

Quadratic equations whose discriminants are equal to zero have one repeated root (solution). Because this root appears twice in the quadratic equation, it has a multiplicity of 2. The number of times a factor appears in a polynomial, such as quadratic, is its multiplicity.

The -intercept of the graph of a function  is the point at which it crosses the -axis; its -coordinate is 0, so its -coordinate is . 

,

so, by setting ,

The -intercept is .

The direction of the concavity of the parabola of the function

is either upward or downward depending entirely on the sign of , the coefficient of . This coefficient, , is negative; the parabola is concave downward.

The parabola in question is a vertical parabola. Its equation is in the standard form

Before the focus can be found, it is necessary to find the vertex . This is located at the point with abscissa

.

Substitute this for to find the ordinate:

The vertex of the parabola is .

The focus of a vertical parabola is located at the point

.

Setting , the point has coordinates

.

, so the focus is at

.