SOLUTION: graph the function f by starting with the graph of y = x2 and using transformations (shifting, compressing, stretching, and/or reflection). [Hint: If necessary, write f in the

Algebra ->  Graphs -> SOLUTION: graph the function f by starting with the graph of y = x2 and using transformations (shifting, compressing, stretching, and/or reflection). [Hint: If necessary, write f in the       Log On


   

Question 466195: graph the function f by starting with the graph of y = x2 and using transformations (shifting, compressing, stretching, and/or reflection).
[Hint: If necessary, write f in the form f(x) = a(x – h)2 + k.]
f(x) = –2x2 + 6x + 2
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
graph the function f by starting with the graph of y = x2 and using transformations (shifting, compressing, stretching, and/or reflection).
[Hint: If necessary, write f in the form f(x) = a(x – h)2 + k.]
f(x) = –2x2 + 6x + 2
..
–2x^2 + 6x + 2
completing the square
-2(x^2-3x+9/4)+2+9/2
-2(x-3/2)^2+13/2
This is a parabola of the standard form, y=A(x-h)^2+k, with h=3/2 and k=13/2, and it opens downward.
see the graph of this parabola as follows:
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+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C-2%28x-3%2F2%29%5E2%2B13%2F2%29+
..
The next graph shows the result of removing the coefficient A=2. Notice how the curve is a little wider or less steep. The larger this coefficient, the steeper the curve.
..
+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+-%28x-3%2F2%29%5E2%2B13%2F2%29+
..
The next graph shows the result of removing (h,k) the (x,y) coordinates of the vertex.
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+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+-x%5E2%29+
..
Removing the negative coefficient
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+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2%29+
..