# Graphing Quadratic Functions - chaoticgolf.com Graphing Quadratic Functions A Shortcut and A Summary All the slides in this presentation are timed. Trying to advance the slides before you are asked to do so will result in skipping part of the presentation on that slide. When each slide is finished a box will appear to let you know there is nothing left on that slide. DONE Before we begin Things you should know/understand before we begin. 1. What is a parabola? 2. What is the vertex of a parabola? 3. How do you find the vertex of a parabola? a) In Standard Form y = ax2 + bx + c b) In Vertex Form y = a (x h)2 + k c) In Intercept Form y = a (x p)(x q) 4. What is the line of symmetry of a parabola? 5. The symbol, , means change. Thus x, means the change in x. DONE A Little Exploration Fill out the following table. Click the screen when you are done. x x y = x2 y 0

0 1 1 1 1 1 2 4 3 1 3 9 5 1 4 16 7 1 5

25 9 1 6 36 11 1 7 49 13 1 8 64 15 DONE A Little Exploration Fill out the following table. Click the screen when you are done. x x

y = 2x2 y 0 0 1 1 2 2 = 2(1) 1 2 8 6 = 2(3) 1 3 18 10 = 2(5) 1 4 32

14 = 2(7) 1 5 50 18 = 2(9) 1 6 72 22 = 2(11) 1 7 98 26 = 2(13) 1 8 128 30 = 2(15) DONE A Little Exploration

Fill out the following table. Click the screen when you are done. x x y = 3x2 y 0 0 1 1 3 3 = 3(1) 1 2 12 9 = 3(3) 1 3 27 15 = 3(5)

1 4 48 21 = 3(7) 1 5 75 27 = 3(9) 1 6 108 33 = 3(11) 1 7 147 39 = 3(13) 1 8

192 45 = 3(15) DONE A Little Exploration Fill out the following table. Click the screen when you are done. x x y = 0.5 x2 y 0 0 1 1 0.5 0.5 = 0.5(1) 1 2 2 1.5 = 0.5(3) 1

3 4.5 2.5 = 0.5(5) 1 4 8 3.5 = 0.5(7) 1 5 12.5 4.5 = 0.5(9) 1 6 18 5.5 = 0.5(11) 1 7 24.5

6.5 = 0.5(13) 1 8 32 7.5 = 0.5(15) DONE A Little Exploration Fill out the following table. Click the screen when you are done. x x y = ax2 y 0 0 1 1 a 1a 1 2

4a 3a 1 3 9a 5a 1 4 16a 7a 1 5 25a 9a 1 6 36a 11a

1 7 49a 13a 1 8 64a 15a DONE So What? First notice that in every example we just did the vertex was at (0, 0). The vertex of a parabola is not always at (0, 0), but the patterns we are going to employ will still work IF you start at the vertex. Second, as we moved right on the x-axis, how far did we go every time? 1 Third, as we moved up the y-axis, we followed a pattern. Did you see that pattern? Up a, 3a, 5a, 7a, 9a, etc. Notice that in every example so far, the a value was positive. That is why we moved UP. If a is negative, then the pattern of odd multiples of a still holds, but you would move DOWN. To put it together, if we want to graph a parabola by hand FIRST: Find the vertex SECOND: To find other points on the parabola without making a table of values, start at the vertex and move right 1, up a then move right 1, up 3a then move right 1 , up 5a etc. THIRD: To finish graphing the parabola, reflect the points you graphed in the last step over the line of symmetry. DONE

Lets Try a Few Graph each parabola below. Click on either the answer to see just the graph or the HELP button to see a step by step explanation. #1: y = 2x2 4x + 1 ANSWER HELP ANSWER HELP ANSWER HELP #2: y = (x 5)2 + 4 #3: y = 3(x + 2)(x + 4) #1: y = 2x2 4x + 1 STANDARD FORM y IVE GOT IT! TRY ANOTHER ONE. x Click one HOW DO YOU DO THIS PROBLEM? #1: y = 2x2 4x + 1

STANDARD FORM First, find the line of symmetry: x= y -b +4 = =1 2a 2(2) Second, find the vertex by plugging in the line of symmetry for the x-value. 2 y = 2 ( 1) - 4 ( 1) +1 = - 1 So, the vertex is (1, 1). x Since a = 2 Move Right 1, Up 2 (1a) Move Right 1, Up 6 (3a) Reflect the points across the line of symmetry and draw the parabola. Click here to go back to problems DONE #2: y = (x 5)2 + 4 VERTEX FORM y

IVE GOT IT! TRY ANOTHER ONE. x Click one HOW DO YOU DO THIS PROBLEM? #2: y = (x 5)2 + 4 VERTEX FORM y First, plot the vertex at (5, 4). Since a = Move Right 1, Down (1a) Move Right 1, Down 3 2 (3a) Move Right 1, Down 5 2 (5a) x The line of symmetry of a parabola is a vertical line through the vertex. Reflect the points across the line of symmetry and draw the parabola. Click here to go back to problems DONE #3: y = 3(x + 2)(x + 4) INTERCEPT FORM y

IVE GOT IT! TRY ANOTHER ONE. x Click one HOW DO YOU DO THIS PROBLEM? #3: y = 3(x + 2)(x + 4) INTERCEPT FORM y First, the x-intercepts are the values of x such that each factor equals 0. So, the x-intercepts are x = 2 and x = 4. The line of symmetry is halfway between these two values. So, the line of symmetry is x = 3. Plug this value of x into the original equation to find the yvalue of the vertex. y =- 3( - 3 + 2) ( - 3 + 4) = 3 x So the vertex is (3, 3) Since a = 3 Move Right 1, Down 3 (1a) Move Right 1, Down 9 (3a) Reflect the points across the line of symmetry and draw the parabola. Click here to go back to problems DONE

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