A vertical line has infinitely many values of  but only one value of . Thus, vertical lines are of the form , where  is a real number. The only equation of this form is . 

To solve for the -intercept, set  to zero and solve for :

To solve for the -intercept, set  to zero and solve for :

Use substution to solve this problem:

becomes and then is substituted into the second equation. Then solve for :

, so and to give the solution .

Slope-intercept form is .

A parabola is one example of a quadratic function, regardless of whether it points upwards or downwards.

The red line represents a quadratic function and will have a formula similar to .

The blue line represents a linear function and will have a formula similar to .

The green line represents an exponential function and will have a formula similar to .

The purple line represents an absolute value function and will have a formula similar to .

A parabola is a curve that can be represented by a quadratic equation.  The only quadratic here is represented by the function , while the others represent straight lines, circles, and other curves.

To find the center or the radius of a circle, first put the equation in the standard form for a circle:  , where  is the radius and  is the center.

From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula  . 

, so .

 and , so  and .

Therefore, .

Because the constant, in this case 4, was not in the original equation, we need to add it to both sides:

Now we do the same for :

We can now find :

To find the center or the radius of a circle, first put the equation in standard form:  , where  is the radius and  is the center.

From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula  . 

, so .

 and , so  and .

This gives .

Because the constant, in this case 9, was not in the original equation, we must add it to both sides:

Now we do the same for :

We can now find the center: (3, -9)

To find the -intercepts (where the graph crosses the -axis), we must set . This gives us the equation:

Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:

 and 

We can then solve these two equations to obtain .

To find the -intercepts (where the graph crosses the -axis), we must set . This gives us the equation:

Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:

 and 

We can then solve these two equations to obtain