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Question 92259: Hello,
As far as I know these are not text book problems. These problems were on an exam and I do not know where to start them how to start them any of that good stuff. Hoping someone can help me.
Here is the first problem...
Identify the axis of symmetry, create a suitable table of values, and then sketch the graph (including the axis of symmetry).
y = –x^2 + 3x – 3
I need to know what to do where to start all of it. At this point I am just soo confused.
Second problem is similar to the first.
Identify the axis of symmetry, create a suitable table of values, and then sketch the graph (including the axis of symmetry).
y = x^2 – 4x
Can anyone help me with these problems. If so I will be soooo greatful... I need all the help I can get, I have one week left in my current math class and at this point I have a failing grade. Please Help.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Identify the axis of symmetry, create a suitable table of values, and then sketch the graph (including the axis of symmetry).
y = –x^2 + 3x – 3
Complete the square on the x-terms as follows:
x^2-3x+? = -(y+3)+?
? = [ (1/2) the coefficient of the 1st degree term]^2 :
? = [(1/2)(-3)]^2 = 9/4
Replace ? with 9/4 to get:
x^2-3x+(9/4) = -y-3+(9/4)
Factor the left side ; simplify the right side to get:
(x-(3/2))^2 = -y-(3/4)
(x-(3/2))^2 = -(y+(3/4))
This form shows you the vertex and axis of symmetry of the parabola
The vertex is ((3/2),(-3/4))
The axis of symmetry is x=(3/2)
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Table of values for y = –x^2 + 3x – 3
If x=0, y=-0^+3*0-3 = -3
if x=1, y=-1^2+3*1-3 = -1
if x=-1, y=-(-1)^2-3*-1-3 = 1
etc.
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Identify the axis of symmetry, create a suitable table of values, and then sketch the graph (including the axis of symmetry).
y = x^2 – 4x
x^2-4x+? = y+?
? = [(1/2)(-4)]^2 = 4
x^2-4x+4 = y+4
(x+2)^2 = y+4
Vertex is (-2,-4)
Axis of symmetry is x=-2
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To generate a table of values choose x-values around x=-2 like -1.-2
0,1; substitute them one at a time into y = x^2 – 4x to find the
corresponding y-values.
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Cheers,
Stan H.
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