Another method of writing an equation in vertex form is to - PowerPoint Presentation

Slide1

Another method of writing an equation in vertex form is to complete the squareIf you have an equation in the form h = -2.25x2 + 4.5x + 6.75, where ‘h’ is height, how do you find the maximum height?1st, factor out -2.25:h = -2.25(x2 - 2x – 3)We need the equation in vertex form, so try finding a perfect square, or make one

6.3 - Completing the SquareSlide2

h = -2.25(x2 - 2x – 3)When figuring out what to factor out, consider the first 2 terms (i.e. x2 – 2x)h = -2.25((x – 1)2 – 4)Now expand the expression:h = -2.25(x – 1)2 – (-2.25)(4)

h = -2.25(x – 1)

2 + 9The vertex is (1, 9), therefore, the maximum height is 9 meters.

6.3 - Completing the SquareSlide3

Write y = x2 + 6x + 2 in vertex form, and then graph the relation.Notice that the equation can’t be factored normally, so let’s try to complete the squarey = x2 + 6x + 2 – let’s try (x + 3)2 because the first 2 terms are x2 + 6x(x + 3)

2

= x2 + 6x + 9, but our last term is +2, so we need to subtract 7 to make the expressions matchy = (x +

3)

2 - 7

Example #2Slide4

y = (x + 3)2 - 7Therefore, the vertex is at (-3, -7).Since a > 0, the parabola opens upwardsThe equation of axis of symmetry is x = -3The y-intercept is 2 (set x = 0)Example #2 cont’dSlide5

Carrie’s diving platform is 6 ft above the water. One of her dives can be modeled by the equation d = x2 – 7x + 6, where d is her position relative to the surface of the water and x is her horizontal distance from the platform. How deep did Carrie go before coming back up to the surface?Example #3Slide6

d = x2 – 7x + 6This looks more strange than the others, but it’s the same. We are dealing with x2 – 7xTry (x – 3.5)2 = x2 – 7x + 12.25We need +6, not +12.25, so we need to subtract 6.25 from the perfect square to make the 2 expressions equald = (x – 3.5)2 – 6.25The vertex is (3.5, 6.25), so Carrie dove to a depth of 6.25

ft

before turning back.

Example #

3 cont’dSlide7

A football’s height h after t seconds is:h = -4.9t2 + 11.76t + 1.4h = -4.9(t2 – 2.4t – 0.29)For (t2 – 2.4), try (t – 1.2)2 (t – 1.2)

2

= t2 – 2.4t + 1.44We need -0.29 not +1.44 so we need to subtract 1.73 from (t – 1.2)

2

to make the expressions equalh = -4.9((t – 1.2)

2

– 1.73)

h = -

4.9(t

– 1.2)

2

–

8.48

Therefore, the football reached a maximum height of 8.48m after 1.2 seconds.

Example

#4Slide8

A quadratic relation in standard form, y = ax2 + bx + c an be rewritten in its equivalent vertex form, y = a(x – h)2 + k, by creating a perfect square within the expression and then factoring itThis technique is called completing the squareIn Summary…