## How do we identify them?

·         The second difference is the same
·         The graph is in the shape of a parabola/ a U.
·         The eq’n has an x2 value.

## Everything you need to know about parabolas.

·         Parabolas can open up or down.
·         The zero of a parabola is where the graph crosses the x–axis.
·         Zero can also be called roots or x-intercept.
·         The axis of symmetry divides the parabola into 2 equal halves.
·         The y-intercept of a parabola is where the graph crosses the y-axis.
·         The optimal value is where the graph is at its maximum or minimum.
·         The vertex is the point where the axis of symmetry and optimal value cross.

## How parabolas move.

Vertical movement

·         y = x2 + 3  à Translates up 3 units
·         y = x2 – 3  à Translates down 3 units

Vertex movment

·         +k à  when k is positive the movment is up
·         –k  à when k is negative the movment is down

Horizontal movment

·         y = (x + 3)2 à translates 3 units left
·         y = (x – 3)2 à translates 3 units right

Compression/Stretch

·         y = 1x2  à vertically compressed by a factor of 1
2                                                                    2
·         y = 2x2 à vertically stretched by a factor if 2

Reflection

·         y =– x2  à reflected in the x – axis
·         when a is positive, it opens upwords.
·         When a is negative, it opens downwards.

y = a(x – h)2 + k

k
(+) up
(-) down

h
(+) right
(-) left

a
(+) opens up
(-)  opens down
(whole #) stretched
(faction) compressed

When a basketball is trown upward, it’s path can be modelled by the function:

y = –4.5(x – 0.1u)2 + 4
When u is velocity(speed)
x is time
y is the height

a. What is the eq’n when the ball is thrown with a velocity of 12m/s?
·         sub 12 into u
y = –4.9 (x – 0.1(12))2 + 9
y = –4.9 (x – 1.2)2 + 9

b. How many seconds after it is thrown does the ball reach the highest point?

Since the vertex is (1.2,9) we know that it reaches the maximum height of 1.2s.