Download Section 3.1 Graphing Quadratic Functions in Vertex Form

Pre‐Calculus 20 Chapter 3 Notes Section 3.1 Quadratic Functions in Vertex Form y
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Using a table of values, graph y = x and y = 2x Table of Values x y=x2 y=2x2 ‐3 9 18 ‐2 4 8 x
‐1 1 2 0 0 0 1 1 2 2 4 8 3 9 18 y
Using a table of values, graph y = ‐1x2 (or y = ‐x2) Table of Values x y=‐x2 y=‐2x2 ‐3 ‐9 ‐18 ‐2 ‐4 ‐8 ‐1 ‐1 ‐2 x
0 0 0 1 ‐1 ‐2 2 ‐4 ‐8 3 ‐9 ‐18 Graphing Quadratic Functions:  An equation (function) with a degree of 2 (highest exponent is 2) is called a “quadratic function”.  y = x2 is the simplest form.  f(x) = x2 is another way to write the equation (function notation). Recall: A function has no values of "x" that repeat. The "y" values can repeat.  The graph’s shape is called a “parabola”.  The “vertex” of the graph is the lowest point (if the graph faces up). We say that this graph then has a “minimum” value for y.  The “vertex” of the graph is the highest point (if the graph faces down). We say that this graph then has a “maximum” value for y.  The axis of symmetry divides a parabola into 2 equal halves.  The general form of a quadratic function can be written as: y = a(x – p)2 + q OR f(x) = a(x – p)2 + q  Notice the 1A, 3A, 5A pattern (key to quickly graphing quadratics!) How the values of “a”, “p” , “q” affect the equation y = a(x – p)2 + q Part 1: The affect “a” has on the quadratic function: y = ax2 Using a table of values, graph all 5 functions onto one grid to compare them. If possible use different colors to really see the differences. a=1 a=2 a=½ a=‐2 a=‐3 Table of Values Table of Values Table of Values Table of Values Table of Values 2
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x y=‐3x2 x y=x x y=‐2x x y=2x x y=½x ‐2 ‐12 ‐2 4 ‐2 ‐8 ‐2 8 ‐2 2 ‐1 ‐3 ‐1 1 ‐1 ‐2 ‐1 2 ‐1 .5 0 0 0 0 0 0 0 0 0 0 1 ‐3 1 1 ‐2 1 2 1 .5 1 2 ‐12 2 4 2 ‐8 2 8 2 2 Summarize what you see: 1. If a > 0, then the graph opens UP. 2. If a < 0, then the graph opens DOWN. {We call this a “REFLECTION IN THE X‐AXIS”.} 3. If |a| > 1 then the graph gets skinnier (thinner). 4. If |a| < 1 then the graph gets wider.
How the values of “a”, “p” , “q” affect the equation y = a(x – p)2 + q Part 2: The affect “q” has on the quadratic function: y = x2 + q Using a table of values, graph all 3 functions onto one grid to compare them. If possible use different colors to really see the differences. q=0 q=3 q=‐2 Table of Values Table of Values Table of Values x y=x2 x y=x2 + 3 x y=x2 – 2 ‐2 4 ‐2 7 ‐2 2 ‐1 1 ‐1 4 ‐1 ‐1 0 0 0 3 0 ‐2 1 1 1 4 1 ‐1 2 4 2 7 2 2 Summarize what you see: 1. The value of “q” shifts the parabola up and down. {We call this a “VERTICAL TRANSLATION UP OR DOWN”.} 2. If q > 0, the graph shifts up. 3. If q < 0, the graph shifts down. How the values of “a”, “p” , “q” affect the equation y = a(x – p)2 + q Part 3: The affect “p” has on the quadratic function: y = (x – p)2 Using a table of values, graph all 3 functions onto one grid to compare them. If possible use different colors to really see the differences. Take note of the p=0 p=2 p=‐3 signs for the Table of Values Table of Values Table of Values values of “p”. x y=x2 x y=(x – 2)2 x y=(x + 3)2 ‐2 4 ‐2 16 ‐7 16 ‐1 1 ‐1 9 ‐6 9 0 0 0 4 ‐5 4 1 1 1 1 ‐4 1 2 4 2 0 ‐3 0 3 1 ‐2 1 4 4 ‐1 4 5 9 0 9 6 16 1 16 Notice that the y values were not altered. This time it was the “x” values that “shifted”. Summarize what you see: 1. The value of “p” shifts the parabola left and right. {We call this a “HORIZONTAL TRANSLATION LEFT OR RIGHT”.} 2. If p > 0, the graph shifts right. 3. If p < 0, the graph shifts left. Using the Mini‐Grids handed out, graph the following by: A. Finding the vertex. B. Graphing using the 1A, 3A, 5A pattern 1. y = x2 – 5 or y = 1(x – 0)2 – 5 2. y = ‐x2 + 6 or y = ‐1(x – 0)2 + 6 3. y = (x – 1)2 or y = 1(x – 1)2 + 0 4. y = (x + 2)2 or y = 1(x + 2)2 + 0 5. y = (x – 2)2 – 5 6. y = ‐(x + 1)2 + 4 7. y = 2(x – 1)2 – 7 8. y = 3x2 – 10 9. y = ‐2(x – 2)2 10. y = ½x2 Finding the Vertex, Domain, Range, Axis of Symmetry, Minimum/Maximum Values “p” is always the Example 1: Graph the function f(x) = 2(x + 6)2 – 3 opposite sign of what is in the brackets. Step1: Find the vertex first. The vertex is always (p, q). Vertex is (‐6, ‐3) (Note the signs!!) “q” keeps the same sign. p = ‐6 q = ‐3 (This is very important for steps 3 to 5) Step2: The domain (“numbers x can be”) for this graph is {x | x  R}. This is the symbolic way to say that x can be any real number. Step3: The axis of symmetry is a vertical line that will cut the graph in half. The equation is always x = p. (All vertical lines are x = ?.) Since p = ‐6, the eqn of the axis of symmetry is x = ‐6. Note: Use the “q” value from the vertex Step4: Does the graph open UP or DOWN? for the minimum It opens UP since the “a” value is positive! value (for y) and for This means we have a Minimum Value (for y) of ‐3. the range. Step5: The range (“numbers y can be”) for this graph is {y | y  R | y ≥ ‐3}. This is the symbolic way to say that y can be any real number that is greater than or equal to ‐3. The reason we use the “greater than” symbol is that we know the graph faces UP so all the y values are starting at ‐3 and getting bigger. Graph using Vertex and Staircase Method (Uses pattern 1a, 3a, 5a, …) y
** From 0, 1, 4, 9, 16, etc pattern Fill in the Vertex and x values surrounding the vertex. Table of Values From the vertex, increase x y the y values by the ‐4 staircase pattern: ‐5 1a = 1(2) = 2 (up 2) ‐6 ‐3 3a = 3(2) = 6 (up 6) ‐7 5a etc. ‐8 x
Table of Values x y ‐4 5 ‐5 ‐1 ‐6 ‐3 ‐7 ‐1 ‐8 5 Plot this table for the graph of y = 2(x +6)2 – 3 “p” is always the opposite sign of what is in the brackets. “q” keeps the same sign. Example 2: Graph the function f(x) = ‐2(x – 6)2 + 3 Step1: Find the vertex first. The vertex is always (p, q). Vertex is (6, 3) (Note the signs!!) p = 6 q = 3 (This is very important for steps 3 to 5) Step2: The domain (“numbers x can be”) for this graph is {x | x  R}. This is the symbolic way to say that x can be any real number. Step3: The axis of symmetry is a vertical line that will cut the graph in half. The equation is always x = p. (All vertical lines are x = ?.) Since p = 6, the eqn of the axis of symmetry is x = 6. Note: Use the “q” value from the vertex Step4: Does the graph open UP or DOWN? for the minimum It opens DOWN since the “a” value is negative! value (for y) and for This means we have a Maximum Value (for y) of 3. the range. Step5: The range (“numbers y can be”) for this graph is {y | y  R | y ≤ 3}. This is the symbolic way to say that y can be any real number that is less than or equal to 3. The reason we use the “less than” symbol is that we know the graph faces DOWN so all the y values are starting at 3 and getting smaller. Graph using Vertex and Staircase Method y
(Uses pattern 1a, 3a, 5a, …) ** From 0, 1, 4, 9, 16, etc pattern Fill in the Vertex and x values surrounding the vertex. Table of Values From the vertex, increase x y the y values by the 4 staircase pattern: 5 1a = 1(‐2) = ‐2 (down 2) 6 3 3a = 3(‐2) = ‐6 (down 6) 7 5a etc. x
8 Table of Values x y 4 ‐5 5 1 6 3 7 1 8 ‐5 Plot this table for the graph of y = ‐2(x – 6)2 + 3 X‐Intercepts, Y‐Intercepts =x‐intercept is where the graph crosses the x‐axis =y‐intercept is where the graph crosses the y‐axis Example 3: Determine the number of x‐intercepts for the following functions: a) f(x) = .5x2 – 7 First find "a" and determine if the graph faces UP or DOWN. a = .5 Faces UP Then find the vertex (p,q). vertex is (0, ‐7) Draw a rough graph to see how many x‐intercepts there are. b) f(x) = ‐2(x + 1)2 a = ‐2, Faces DOWN, vertex is (‐1, 0) 1
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c) f(x) =  x  5  11 6
a =  1 , Faces DOWN, vertex is (5, ‐11) 6
Extra Practice Questions: Graph the following: 2. y = ‐2(x – 3)2 3. y = 3(x + 2)2 – 6 1. y = x2 + 1 Example4: Given the quadratic functions below find the x and y intercepts: You may round off your answers to the nearest tenth if needed. a) f(x) = .5x2 – 8 To find y‐intercepts, substitute x = 0 and find y. y = .5x2 – 8 {Notice how we changed f(x) to “y” to help us solve for y.} y = .5(0)2 – 8 y = ‐8 So the y‐intercept is ‐8 {There will always be a y‐intercept for quadratic functions.} To find x‐intercepts, substitute y = 0 and solve for x. y = .5x2 – 8 0 = .5x2 – 8 8 = .5x2 16 = x2 {div both sides by .5} x =  16 {don’t forget the ± symbol since we have 2 possible answers} x = ±4 So there are two x‐intercepts {‐4 and 4} b) f(x) = (x – 3)2 – 2 To find y‐intercepts, substitute x = 0 and find y. y = (x – 3)2 – 2 y = (0 – 3)2 – 2 y = 9 – 2 y = 7 So the y‐intercept is 7 To find x‐intercepts, substitute y = 0 and solve for x. y = (x – 3)2 – 2 0 = (x – 3)2 – 2 2 = (x – 3)2  2 = x – 3 {square root both sides} x = 3  2 So there are two x‐intercepts {1.6 and 4.4} Textbook Assignment: Page 157‐160 #1ac, 2all, 4ac, 6all Example 5: Suppose a parabolic archway has a width of 280cm and a height of 216cm at its highest point above the floor. a) Write a quadratic function in vertex form that models the shape of this archway. An equation is written in the form y = a(x – p)2 + q We need to find values for a, p and q. First, the vertex is (0, 216) so p = 0, q = 216. To find "a", we need the vertex and another point on the curve. We can use the point (140, 0). We could also use the point (‐140, 0) to get the same result. So, p = 0, q = 216 and x = 140 when y = 0. Take note that this 2
looks just like we did y = a(x – p) + q 2
when finding the x‐
0 = a(140 – 0) + 216 The exact value of "a" is" 2
intercepts. Why? ‐216 = a(140) 216
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‐0.011 = a The equation for the function is y = ‐.011(x – 0)2 + 216 {a = ‐.011, p = 0, q = 216} OR y = ‐.011x2 + 216 b) Determine the height of the archway at a point that is 50 cm from its outer edge.
If we substitute x = 90 or x = ‐90, we can find the value for y in our equation. y = ‐.011(x)2 + 216 y = ‐.011(90)2 + 216 y = ‐89.1 + 216 y = 126.9 cm (Note: Using the exact fraction for a, we get y = 126.7 cm) The archway is 126.9 cm at a point 50 cm from the outer edge. Example 6: (Do #10 in textbook, pg 158, like this!!) Rewrite the new equation of f(x) = x2 if the graph is translated 3 units to the left, 4 units up and reflected across the x‐axis. p = ‐3 q = 4 Solution: f(x) = 1(x + 3)2 + 4 left 3 and 4 up f(x) = ‐1(x + 3)2 – 4 reflected across x‐axis means a is now ‐1, but also q is now ‐4 Textbook Assignment: Page 158‐160 #3ab, 7all, 8all, 9ab, 10all, 16all, 17