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Question 674318: For each equation, identify the equation of the axis of symmetry, the coordinates of the vertex, the y-intercept, the zeros, and the maximum or minimum values.
y = x2 + 4x + 3
y = -2(x - 3)2 + 9
y = 3(x - 0.5)2
help??
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! For each equation, identify the equation of the axis of symmetry, the coordinates of the vertex, the y-intercept, the zeros, and the maximum or minimum values.
y = x2 + 4x + 3
y = -2(x - 3)2 + 9
y = 3(x - 0.5)2
**
rewriting equations with standard notation:
y = x^2 + 4x + 3
y = -2(x - 3)^2 + 9
y = 3(x - 0.5)^2
..
y = x^2 + 4x + 3
complete the square
y=(x^2+4x+4)+3-4
y=(x+2)^2-1
This is an equation of a parabola that opens upwards. Function has a minimum.
Its standard form: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of vertex)
For given equation: y=(x+2)^2-1
vertex: (-2,-1)
axis of symmetry: x=-2
minimum: -1
..
y-intercept
set x=0
y=3
..
zeros:
set y=0
(x+2)^2-1=0
(x+2)^2=1
(x+2)=±√1=±1
x=-2±1
x=-3,-1
..
For given equation:y = -2(x - 3)^2 + 9
This is an equation of a parabola that opens downwards. Function has a maximum.
Its standard form: y=-A(x-h)^2+k, (h,k)=(x,y) coordinates of vertex)
vertex: (3,9)
axis of symmetry: x=3
maximum: 9
..
y-intercept
set x=0
y=-2(-3)^2+9
y=-9
..
zeros:
set y=0
-2(x-3)^2+9=0
(x-3)^2 =-9/-2=9/2
x-3=√9/√2=3/√2=3√2/2≈2.12
x=3±2.12
x=.88,5.12
..
For given equation:y = 3(x - 0.5)^2
This is an equation of a parabola that opens upwards. Function has a minimum.
Its standard form: y=(x-h)^2+k, (h,k)=(x,y) coordinates of vertex)
I will let you this one
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