2.03  The Parabola

College Algebra: One Step at a Time,  Page 219 - 228:   #37, 43, 46, 47, 68, 71, 72, 75

Dr. Robert J. Rapalje

Seminole State College of Florida

Sanford, FL  32773

General Explanation

There are quite a number of different explanations and presentations that can be made for graphing parabolas.  In the analytic geometry explanation, usually found in a Precalculus or Calculus course, the parabola is defined to be the set of all points that are equally distant from a given point called the focal point and a given line, which is called the directrix of the parabola.

By contrast, and for simplicity in lower math courses, a parabola can be explained in terms of simple graphs of equations in the form of

or .   

In the more general case, equations in the form

 and

can be shown to represent graphs of parabolas.  

For , the parabola will open  up  if  a is positive or down if a is negative. 

For , the parabola will open to the right if a is positive or left if a is negative. 

There are several different ways to explain and graph parabolas in this form.  You can just make a table of values and plot points.  I don’t recommend this—it’s too tedious and mechanical.  You can also graph by finding an intercept (the easiest and a very important point to graph!) and the vertex of the parabola by completing the square.  For the past 15 years or so, in my book College Algebra: One Step at a Time, I explained parabolas by this method.  However, I think it might be easier to find the vertex of a parabola by using a simple formula that is easy to derive and remember. 

For a parabola in the form , the vertex will have as its x coordinate  .  This is not a new formula.  It actually comes from the quadratic formula , where the radical is zero.   After you find the x-coordinate of the vertex, then you just substitute this value of  x into the original equation, and solve for y to find the y-coordinate of the vertex.  Remember, the vertex consists of TWO coordinates, both x and y.  You must find BOTH of them. 

It is also a good idea to find the y-intercept.  This is both an easy point and an important point to graph, since it indicates where the graph crosses the y-axis.  It is also very easy to find, since you just let x=0, and solve for y in the original equation.

For an equation of a parabola  in the form, the vertex will have as its y coordinate .  Then substitute this value of y back into the original equation to find the x coordinate.  Don’t forget, you must find BOTH coordinates, and when you give ordered pair notation (x, y), you must always put the x coordinate first.

It is also a good idea to find the x-intercept.  This is both an easy point and an important point to graph, since it indicates where the graph crosses the x-axis.  It is also very easy to find, since you just let y=0, and solve for x in the original equation.

 37.      

Solution:  Before you ever start the problem, you know from the fact that this is an equation in the form of , so it is a PARABOLA, and it opens to the UP!  Knowing this it should be very easy to graph, if you can find the VERTEX and the y-intercept. 

Of course, the easiest point to find is the y-intercept, which is where , and therefore .   This is the point

There are many ways to find the vertex, but perhaps the easiest way is to know that the vertex will be at , where and  .  Therefore

To find the y-coordinate, substitute  back into the original equation for y.

Therefore the vertex is at .

Sketch the graph by locating these two points,   and , and draw a parabola from the vertex at , opening UP, and passing through .

 43.      

Solution:  Before you ever start the problem, you know from the fact that this is an equation in the form of , so it is a PARABOLA, and it opens to the DOWN!  Knowing this it should be very easy to graph, if you can find the VERTEX and the y-intercept. 

Of course, the easiest point to find is the y-intercept, which is where , and therefore .   This is the point .

There are many ways to find the vertex, but perhaps the easiest way is to know that the vertex will be at , where and  .  Therefore

To find the y-coordinate, substitute  back into the original equation for y.

Therefore the vertex is at .

Sketch the graph by locating these two points,   and , and draw a parabola from the vertex at , opening DOWN, and passing through .

46.      

Solution:   Before you ever start the problem, you know from the fact that this is an equation in the form of , so it is a PARABOLA, and it opens to the UP!  Knowing this it should be very easy to graph, if you can find the VERTEX and the y-intercept. 

Of course, the easiest point to find is the y-intercept, which is where , and therefore .   This is the point .

There are many ways to find the vertex, but perhaps the easiest way is to know that the vertex will be at , where and  .  Therefore

To find the y-coordinate, substitute  back into the original equation for y.

Therefore the vertex is at .

Sketch the graph by locating these two points,   and , and draw a parabola from the vertex at , opening UP, and passing through .

47.      

Solution:  Before you ever start the problem, you know from the fact that this is an equation in the form of , so it is a PARABOLA, and it opens to the UP!  Knowing this it should be very easy to graph, if you can find the VERTEX and the y-intercept. 

Of course, the easiest point to find is the y-intercept, which is where , and therefore .   This is the point

There are many ways to find the vertex, but perhaps the easiest way is to know that the vertex will be at , where and  .  Therefore

To find the y-coordinate, substitute  back into the original equation for y.

Therefore the vertex is at .

Sketch the graph by locating these two points,   and , and draw a parabola from the vertex at , opening UP, and passing through .

 68.      

Solution:  Before you ever start the problem, you know from the fact that this is an equation in the form of , so it is a PARABOLA, and it opens to the RIGHT!  Knowing this it should be very easy to graph, if you can find the VERTEX and the x-intercept. 

Of course, the easiest point to find is the x-intercept, which is where , and therefore .   This is the point

There are many ways to find the vertex, but perhaps the easiest way is to know that the vertex will be at , where and  .  Therefore

To find the x-coordinate, substitute  back into the original equation for x.

Therefore the vertex is at .

Sketch the graph by locating these two points,   and , and draw a parabola from the vertex at , opening LEFT, and passing through .

ALTERNATE SOLUTION--Completing the Square Method

68.      

Solution:  As before, you know from the fact that this is an equation in the form of , so it is a PARABOLA, and it opens to the RIGHT!  Knowing this it should be very easy to graph, if you can find the VERTEX and the x-intercept. 

Of course, the easiest point to find is the x-intercept, which is where , and therefore .   This is the point

In order to find the vertex by the completing the square method, you must have the coefficient of to be  by factoring out the , even if fractions result:

You must complete the square to fill in the blank space within the parentheses by taking half of the , which is , and squaring in order to obtain  .  Then you placed a   in the parentheses on the right side of the equation, but you that  was actually multiplied times the  that was outside the parentheses.  Therefore, you must add  , or  to the left side of the equation.  It should look like this: 

Next subtract  from each side:

From this form of the equation, you can see that the vertex (i.e., what zeros out the x term and what zeros out the y term) is   and . 

Therefore the vertex is at .

Sketch the graph by locating these two points,   and , and draw a parabola from the vertex at , opening LEFT, and passing through .

71.               

Solution:    You know from the fact that this is an equation in the form of , so it is a PARABOLA, and it opens to the LEFT!  Knowing this it should be very easy to graph, if you can find the VERTEX and the x-intercept.  Of course, the easiest point to find is the x-intercept, which is where , and therefore .   This is the point

There are many ways to find the vertex, but perhaps the easiest way is to know that the vertex will be at , where and  .  Therefore

To find the x-coordinate, substitute  back into the original equation for x.

Therefore the vertex is at .

Sketch the graph by locating these two points,   and , and draw a parabola from the vertex at , opening LEFT, and passing through .

72.        

Solution:  You know from the fact that this is an equation in the form of , so it is a PARABOLA, and it opens to the RIGHT!  Knowing this it should be very easy to graph, if you can find the VERTEX and the x-intercept. 

Of course, the easiest point to find is the x-intercept, which is where , and therefore .   This is the point

There are many ways to find the vertex, but perhaps the easiest way is to know that the vertex will be at , where and  .  Therefore

To find the x-coordinate, substitute  back into the original equation for x.

Therefore the vertex is at .

Sketch the graph by locating these two points,   and , and draw a parabola from the vertex at , opening RIGHT, and passing through .

75.       

Solution: You know from the fact that this is an equation in the form of , so it is a PARABOLA, and it opens to the RIGHT!  Knowing this it should be very easy to graph, if you can find the VERTEX and the x-intercept. 

Of course, the easiest point to find is the x-intercept, which is where , and therefore .   This is the point

There are many ways to find the vertex, but perhaps the easiest way is to know that the vertex will be at , where and  .  Therefore

To find the x-coordinate, substitute  back into the original equation for x.

Therefore the vertex is at .

Sketch the graph by locating these two points,   and , and draw a parabola from the vertex at , opening RIGHT, and passing through .

        ALTERNATE SOLUTION--Completing the Square Method

75.  .      

Solution:  You know from the fact that this is an equation in the form of , so it is a PARABOLA, and it opens to the RIGHT!  Knowing this it should be very easy to graph, if you can find the VERTEX and the x-intercept. 

Of course, the easiest point to find is the x-intercept, which is where , and therefore .   This is the point

To find the vertex by the completing the square method, you will need to factor out the coefficient of y, which is , from the  and terms.   It should look like this:

 .

Now complete the square by taking half of the , which is , and square, which is .  You must place the  on the blank line within the parentheses on the right side of the equation.  Notice that this  is actually multiplied by the  that is outside the parentheses, so you actually multiplied both sides by  which would be added to the left side.  It looks like this:

Now, subtract from each side of the equation:

The vertex can be easily found now by “zeroing out” the x and y terms, which gives you

  and

Therefore the vertex is at .

Sketch the graph by locating these two points,   and , and draw a parabola from the vertex at , opening RIGHT, and passing through .

Return to main page        Math in Living C O L O R !!

     Return to Intermediate Algebra page  

 Return to College Algebra page