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course MTH 163
Oct 14, 2012 @ 5:59PM
Sketch a graph of y = x^2, from x = -3 to x = 3. Then sketch a graph of y = 2 x^2 over the same domain. •By what factor do we vertically stretch the first graph to obtain the second?
• - Factor of 2.
• What are the three basic points of the first graph (i.e., the vertex and the points 1 unit to the right and left of the vertex)?
• - Vertex = (0,0) LeftV = (-3, 9) RightV = (3, 9)
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The basic points are those for which x = -1, 0 and 1.
You can of course graph the function with the points you list, but they are not what we mean by the 'basic points' of this graph.
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• What are the three basic points of the second graph?
• - Vertex = (0,0) LeftV = (-3, 18) RightV = (3, 18)
Sketch the second graph shifted 1.75 units in the x direction and .75 units in the y direction.
What are the three basic points of this graph?
- Vertex = (1.75, 0.75) LeftV = (-1.25, 18.75) RightV = (4.75, 18.75)
If f(x) = x^2, then what are
• f(x- 1.75), f(x) + .75, 2 f(x) and 2 f(x- 1.75) + .75?
f(x) = x^2 f(x) = (x - 1.75)^2 (x - 1.75)(x - 1.75) x^2 - 3.5x + 3.0625
f(x) = x^2 f(x) = x^2 + .75
f(x) = x^2 f(x) = 2(x^2)
f(x) = x^2 f(x) = 2(x^2 - 3.5x + 3.0625) (2x^2 - 7x + 6.125) + .75 2x^2 - 7x + 6.875
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Good work exanding the square.
However the form
y = 2 ( x - 1.75 )^2
is the simpler version, more closely connected to the shifts and stretches, and would be the preferred standard form in this context.
If you had y = 2x^2 - 7x + 6.875 and wanted to graph it, you would need to return it to the form y = 2 ( x - 1.75 ) ^2.
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&#This looks good. See my notes. Let me know if you have any questions. &#