Chapter 4 Quadratics 4.3 Using Technology to Investigate Transformations

Chapter 4 Quadratics4.3 Using Technology to Investigate

Transformations

Humour Break

4.3 Using Technology to Investigate Transformations

• The relation y = x² is the simplest quadratic relation. It is the base curve for all relations


• y = x²... a = 1 and the graph opens up, standard width

• This equation is both in vertex form and in standard form

• Consider... y = 1x² + 0x + 0 (standard form)• Consider... Y = a(x – h)² + k• Consider... y = 1(x – 0)² + 0 (vertex form)• Vertex is (0, 0) and y-intercept is also (0, 0)


x y-3 9-2 4-1 10 01 12 43 9

y = x²


• The relation y = -x² is the simplest quadratic relation reflected down about the x axis.


• y = -x²... a = -1 and the graph opens down, standard width


• Consider... y = -1x² + 0x + 0 (standard form)• Consider... y = a(x – h)² + k• Consider... y = -1(x – 0)² + 0 (vertex form)• Vertex is (0, 0) and y-intercept is also (0, 0)


x y-3 -9-2 -4-1 -10 01 -12 -43 -9

y = - x²


• Consider…• y = 2x² and y = -2x²• What impact does the 2 have?


• y = 2x²... a = 2 and the graph opens up, more narrow width (double height for any given point)


• Consider... y = 2x² + 0x + 0 (standard form)• Consider... y = a(x – h)² + k• Consider... y = 2(x – 0)² + 0 (vertex form)• Vertex is (0, 0) and y-intercept is also (0, 0)


x y-3 18-2 8-1 20 01 22 83 18

y = 2x²


• y = -2x²... a = -2 and the graph opens down, more narrow width (double height for any given point)


• Consider... y = -2x² + 0x + 0 (standard form)• Consider... y = a(x – h)² + k• Consider... y = -2(x – 0)² + 0 (vertex form)• Vertex is (0, 0) and y-intercept is also (0, 0)


x y-3 -18-2 -8-1 -20 01 -22 -83 -18

y = -2x²


• So… • y = 2x² and y = -2x²• an “a” of 2 doubles the height of the graph for

a given x value• an “a” of -2 doubles the height of the graph

for a given x value but opening down• We can generalize this rule for different values

of “a”


• Consider…• y = 1/2x² and y = - 1/2x² • What impact does the a of ½ and -½ have?


• y = 1/2x²... a = 1/2 and the graph opens up, but wider (half height for any given point)


• Consider... y = 1/2x² + 0x + 0 (standard form)• Consider... y = a(x – h)² + k• Consider... y = 1/2(x – 0)² + 0 (vertex form)• Vertex is (0, 0) and y-intercept is also (0, 0)


x y-3 4.5-2 2-1 0.50 01 0.52 23 4.5

y = 1/2x²


• y = -1/2x²... a = -1/2 and the graph opens down, but wider (half height for any given point)


• Consider... y = -1/2x² + 0x + 0 (standard form)• Consider... y = a(x – h)² + k• Consider... y = -1/2(x – 0)² + 0 (vertex form)• Vertex is (0, 0) and y-intercept is also (0, 0)


x y-3 -4.5-2 -2-1 -0.50 01 -0.52 -23 -4.5

y = -1/2x²


• So… • y = 1/2x² and y = - 1/2x² • an “a” of ½ halves the height of the graph for a

given x value• an “a” of -1/2 halves the height of the graph

for a given x value but opening down• We can generalize this rule for different values

of “a”


• Consider…• y = x² + 1 and y = x² - 1 • What impact does adding or subtracting 1

have to the graph?


• y = 1x² + 1... a = 1 and the graph opens up• This equation is both in vertex form and in

standard form• Consider... y = x² + 0x + 1 (standard form)• Consider... y = a(x – h)² + k• Consider... y = 1(x – 0)² + 1 (vertex form)• Vertex is (0, 1) and y-intercept is also (0, 1)


x y-3 10-2 5-1 20 01 22 53 10

y = x² + 1


• y = 1x² - 1... a = 1 and the graph opens up• This equation is both in vertex form and in

standard form• Consider... y = x² + 0x - 1 (standard form)• Consider... y = a(x – h)² + k• Consider... y = 1(x – 0)² - 1 (vertex form)• Vertex is (0, -1) and y-intercept is also (0, -1)


x y-3 8-2 3-1 00 -11 02 33 8

y = x² - 1


• So…• y = x² + 1 and y = x² - 1 • a “k” of +1 outside the brackets shifts the

entire graph up by 1• a “k” of -1 outside the brackets shifts the

entire graph down by 1• We can generalize this rule for different values

of “k”


• Consider…• y = (x - 1)² and y = (x + 1)² • What impact does adding or subtracting 1

inside the brackets have to the graph?


• y = 1(x - 1)²... a = 1 and the graph opens up• This equation is both in vertex form and in

standard form• Consider... y = x² - 2x + 1 (standard form)• Consider... y = a(x – h)² + k• Consider... y = 1(x – 1)² + 0 (vertex form)• Vertex is (1, 0) and y-intercept is also (0, 1)


x y-3 16-2 9-1 40 11 02 13 4

y = (x - 1)²


• y = 1(x – (- 1))² or y = 1(x + 1)² ... a = 1 and the graph opens up


• Consider... y = x² + 2x + 1 (standard form)• Consider... y = a(x – h)² + k• Consider... y = 1(x + 1)² + 0 (vertex form)• Vertex is (-1, 0) and y-intercept is also (0, 1)


x y-3 4-2 1-1 00 11 42 93 16

y = (x + 1)²


• So…• y = (x - 1)² and y = (x + 1)² • a “h” of -1 in the brackets (with the subtraction providing

the negative) shifts the entire graph to the right by 1• a “h” of +1 in the brackets (with the double negative

providing the positive) shifts the entire graph to the left by 1

• We can generalize this rule for different values of “h”• Unlike k, the general rule shift is counter-intuitive

because you move in the opposite direction of the sign


• y = a(x – h)² + k… putting it together • “a” opens up & “-a” opens down• If a>1, the graph is more narrow & higher by a

factor of “a”, so if a = 2, the y value for a given x will be twice as high

• If a<1, the graph is wider & flatter by a factor of “a”, so if a = 1/2, the y value for a given x will be ½ as high


• y = a(x – h)² + k… putting it together• (x – h) moves the x of the vertex from 0 right by

“h” so x of vertex of (x – 2) would be at x=2• (x + h) moves the x of the vertex from 0 left by

“h” so x of vertex of (x + 2) would be at x=-2• + k moves the vertex (and graph) up by k• - k moves the vertex (and graph) down by k


• Ex. 1 Write the relation for a parabola that satisfies each of the following conditions:

• Vertex at (4,7), opens downward, same shape as y = x²


• Ex. 1 Write the relation for a parabola that satisfies each of the following conditions:

• Vertex at (4,7), opens downward, same shape as y = x²

• y = - 1(x – 4)² + 7


y = x² andy = -1(x – 4)² + 7


• Ex. 2 Consider a parabola that is congruent to y = x², opens up and has the vertex at (1, -3). Now find the equation of a new parabola, that results if P is:

• (a) stretched vertically by a factor of 3• (b) compressed by a factor of 3• (c) translated 2 units to the left• (d) translated 3 units up• (e) reflected about the x-axis and translated 2 units to

the left and 4 units down and stretched by a factor of 2




• (a) stretched vertically by a factor of 3: • Starting point is y= 1(x -1)² - 3• Stretching vertically by a factor of 3 makes “a” 3• y= 3(x -1)² - 3 (transformed equation)




• (b) compressed by a factor of 3: • Starting point is y= 1(x -1)² - 3• Compressing by a factor of 3 makes “a” 1/3• y= 1/3(x -1)² - 3 (transformed equation)




• (c) translated 2 units to the left: • Starting point is y= 1(x -1)² - 3• Translating 2 units to the left, moves the x of the

vertex from 1 to -1, which changes the (x -1) to (x- (-1)) which is the same as (x + 1)• y= 1(x + 1)² - 3 (transformed equation)




• (c) translated 3 units up: • Starting point is y= 1(x -1)² - 3• Translating 3 units up adds 3 to the k of -3• y= 1(x + 1)² (transformed equation)




• (e) reflected about the x-axis and translated 2 units to the left and 4 units down and stretched by a factor of 2

• Starting point is y= 1(x -1)² - 3• y= -2(x + 1)²-7 (transformed equation)


Homework

• Tuesday, May 17th, p.363, #1-4, 6-13 & 16

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Chapter 4 Quadratics 4.3 Using Technology to Investigate Transformations