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Graphing Polynomials
Rearrange the Equation
• 2y – 2x2 – 12x= 18
• Add 4x2 to each side
• Add 12x to each side
• 2y= 2x2 + 12x + 18
• Divide by 2
• y= x2 + 6x + 9
Find theVertex
• With equation in y= Ax2 + Bx + C form, x-coordinate =-B/2A
• X-coordinate= -6/2 = -3
• Plug -1/3 into equation to find y
• (-3)2 + 6(-3) + 9 = 9 – 18 + 9 = 0
Find Intercepts
• X intercept is found when y=0 (happens to be the vertex)
• Often you will have to factor, and get 2 results.
• In this case, y=(x+3)2
• Y intercept is found when x=0
• Y= (0)2 + 6(0) + 9 = 9
Find Other Easy Points
• Try x=1, x= -1, for instance
• Or 1 to either side of the vertex
• In this case, x= -2, x=-4
Connect Points with Curve
Check End Behavior
• Check to see if the graph looks how it should
• Even powered functions will have their ends pointing the same direction
• If the variable in front of the x2 is positive they will both point up
• If the variable in front of the x2 is negative they will both point down
• We see that this is consistent with our graph
Now try these on your own
• x-y= 2-x2
• x2 – 4 = y - 3x
• 6 – x=x2 + y
Here’s a video to help explain more
Recap
• Rearrange the equation so that y is isolated on one side
• Find the vertex (-b/2a)
• Graph the intercepts [ (0,y) (x,0)]
• Find a few other points
• Draw smooth curve
• Check with end behavior

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Polynomial Graphing Lesson

  • 2. Rearrange the Equation • 2y – 2x2 – 12x= 18 • Add 4x2 to each side • Add 12x to each side • 2y= 2x2 + 12x + 18 • Divide by 2 • y= x2 + 6x + 9
  • 3. Find theVertex • With equation in y= Ax2 + Bx + C form, x-coordinate =-B/2A • X-coordinate= -6/2 = -3 • Plug -1/3 into equation to find y • (-3)2 + 6(-3) + 9 = 9 – 18 + 9 = 0
  • 4. Find Intercepts • X intercept is found when y=0 (happens to be the vertex) • Often you will have to factor, and get 2 results. • In this case, y=(x+3)2 • Y intercept is found when x=0 • Y= (0)2 + 6(0) + 9 = 9
  • 5. Find Other Easy Points • Try x=1, x= -1, for instance • Or 1 to either side of the vertex • In this case, x= -2, x=-4
  • 7. Check End Behavior • Check to see if the graph looks how it should • Even powered functions will have their ends pointing the same direction • If the variable in front of the x2 is positive they will both point up • If the variable in front of the x2 is negative they will both point down • We see that this is consistent with our graph
  • 8. Now try these on your own • x-y= 2-x2 • x2 – 4 = y - 3x • 6 – x=x2 + y
  • 9. Here’s a video to help explain more
  • 10. Recap • Rearrange the equation so that y is isolated on one side • Find the vertex (-b/2a) • Graph the intercepts [ (0,y) (x,0)] • Find a few other points • Draw smooth curve • Check with end behavior

Editor's Notes

  1. I am going to show you all a sample lesson, so that you can see the way I teach.
  2. The ultimate goal is to get Y by itself on one side. Here’s how we’ll do that.
  3. There is another way to find the vertex, if the equation is in a different form, but we’ll stick with this.
  4. The x and y intercepts are easy to plot on the graph and can begin to help us understand what the graph will look like.
  5. Points A and B are the x and y intercepts respectively, and points C and D are the points on either side of the vertex.
  6. Obviously a hand drawn graph will not look perfect, but I should still be able to tell what shape it is, and a few key points will help you do that.
  7. The variable in front of the x2 is called the leading coefficient. It is positive, so our graph should point upwards, which it indeed does.
  8. You will have 10 minutes to try these all on your own. If you get stuck, you can ask me for help. Don’t worry if you don’t get them all, we will go over them.
  9. I’ll show this to the class, though I would skip parts of it. It doesn’t really start until over a minute into the video, and not all of it is necessary to show.
  10. Are there any questions over any of the material? If not, I will assign homework.