The parabola is concave upward if and only if , and concave downward if and only if . Therefore, we need to know the sign of  to answer the question. We show that the two statements together provide insufficient information by examining two equations.

Case 1: 

; we check the values of the expressions in both statements:

The conditions of both statements are satisfied; since , the parabola that graphs this function is concave upward.

Case 2: 

; we check the values of the expressions in both statements:

The conditions of both statements are satisfied; since , the parabola that graphs this function is concave downward.

The two statements together give the vertex of the parabola, but no clue is given as to the orientation of the parabola - horizontal, vertical, or otherwise. Without that information, or any way to find it, the line of symmetry cannot be determined.

Assume both statements are true.

The parabola is concave upward if and only if , and concave downward if and only if . Therefore, we need to know the sign of  to answer the question. We show that the two statements together provide insufficient information by examining two equations.

Case 1: 

If we substitute 0 for ,  is easily seen to be equal to 6, so the -intercept is . Also:

Case 2: 

If we substitute 0 for ,  is easily seen to be equal to 6, so the -intercept is . Also:

Each equation satisfies the conditions of both statements, but  is positive in one equation and negative in the other.  can assume either sign, so the question of the parabola's concavity is unresolved.

Statement 1): The slope is 4.

Write the slope-intercept form, and substitute the slope.

The point and the y-intercept are unknown.  Either of these will be needed to solve for the graph of this line.

Statement 1) by itself is not sufficient to graph a line.

Statement 2): The y-intercept is 4.

Substitute the y-intercept into the incomplete formula.

The function  can then be graphed on the x-y coordinate plane.

Therefore:

Find the graph of .

I)  is a linear equation which passes through the point .

II)  crosses the y-axis at 1300.

To graph a linear equation, we need some combination of slope, y-intercept, or two points.

Statement I tells us  is linear and gives us one point.

Statement II gives us the y-intercept of .

We can use Statement I and Statement II to find the slope of . Then, we can plot the given points and continue the line in either direction to get our graph.

Slope:

Plugging in the provided value of , 1300, we have the equation of the line :

Find the graph of the linear function .

I)  passes through the points  and .

II)  intercepts the -axis at .

Using I), we can find the slope of the function, and then we can start at either point and extend the slope in either direction to find our graph:

So, using I) we are able to find the slope, from which we can find our graph

II) gives us one point, but without any more information, we cannot use II) by itself to find the rest of the graph

So:

Statement I is sufficient to answer the question, but statement II is not sufficient to answer the question.

You cannot determine the equation of a circle knowing only the center, as in Statement 1. 

But given the coordinates of any diameter of a circle, as in Statement 2, you can use the midpoint formula to find the center, and the distance formula to find the distance from the center to either endpoint - this is the radius. Once you know the center and the radius, just apply them to the standard form of the equation of a circle.

The center of the circle of the equation  is the point .

If Statement 2 is assumed, then ; since  is known to be positive,  is negative, which is the same as Statement 1.

From either statement, we know  to be negative, and, subsequently, that both coordinates of the center  are negative, putting it in Quadrant III. 

The equation of a circle is given by:

Where r is the radius and (h,k) are the center of the circle

I) gives us (h,k)

By using II) and distance formula, we can find r

Thus, both statements are needed!

To write the equation of a circle we need the coordinates of its center and the length of its radius.

I) Gives us the center.

II) Tells us that the circle just touches the x-axis at the point (7,0), which is exactly 7 units below our center. This makes our radius 7.

Thus, we need both statements.