Yahoo Answers 06-16-14
- Author:
- Michael Brown
- Topic:
- Parabola
The vertex form of quadratic equation is given by y=a(x-h)^2+k where a, h and are constants and a not equal to 0.
Discuss the change of the shape and position of the graph if
(a) a varies from -2 to 2 when h=3 and k=4.
(b) h varies from -2 to 2 when a=2 and k=4
(c) k varies from -2 to 2 when a=2 and h=4
The vertex form comes from solving the standard form by completing the square.
So you are going from y=Ax^2+Bx+C to y=a(x-h)^2+k
y=4x^2-8x+1
factor out 4
y=4(x^2-2x+___)+1
1/2 of -2 is -1, squared is 1
we are adding 1 inside the parentheses which is really 4 since we are multiplying by 4. to keep the equation the same we have to subtract 4
y=4(x^2-2x+1)+1-4
y=4(x^2-2x+1)-3
y=4(x-1)^2-3
a=4, h=1, k=-3
the minus sign in the binomial is part of the formula
so if it is x-h, h is positive, if it's x+h, h is negative
k behaves normally. +k is positive, -k is negative
the vertex is at (h,k)
a is a scalar
if |a|>1 the parabola is more narrow as a gets bigger
if |a|<1 the parabola is flatter as a gets smaller
if a is positive the parabola opens upward
if a is negative the parabola opens downward
(if a=0 you have a horizontal line y=k)
if a>0 and k>0, the parabola has no real roots (it opens upward and the vertex is above the x-axis, it will not cross the x-axis)
or if a<0 and k<0 it has no real roots (it opens downward and the vertex is below the x-axis)
if k=0 the vertex is on the x-axis and it has one root
if a and k have different signs, the parabola has two real roots
To answer the questions,
a) as a goes from -2 to 2...
the vertex is fixed at (3,4)
it starts as a narrower parabola pointing downward, it will get wider and wider until a=0 and it's a horizontal line then the parabola will open pointing upward and get narrower and narrower
b) as h goes from -2 to 2
this is an upward facing parabola (a bit narrow) since a=2
the vertex will start at (-2,4) and move to the RIGHT until it stops at (2,4)
NOTE: the equation will start at y=2(x+2)^2+4 and stop at y=2(x-2)^2+4
c) as k goes from -2 to 2
again it is narrow and upward facing
the vertex will start at (2,-2) and move UP until it stops at (2,2)