Given 4 coordinate points A(2, 0), B(4, 0), C(5, -2), D(1, -2). If the image is rotated by 180° about the origin O, then C' =
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<h3>Answer:</h3>
- g(x) has an axis of symmetry at x = 3
- g(x) is shifted right 3 units from the graph of f(x)
- g(x) is shifted up 4 units from the graph of f(x)
The vertex form of g(x) is ...
... g(x) = -(x -3)² +4
This is offset to the right by 3 and up by 4 from the parent function. (It is also first reflected across the x-axis.)
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<em>Vertex form</em>
You know the leading coefficient is -1 because that's what it is for x² in the given form. When you factor -1 from the first two terms, of the given form, you have ...
... g(x) = -1(x² -6x) -5
Half the x coefficient inside parentheses will be the constant in the squared binomial term, so that term is (x -3)². The constant in that square is +9, so adding that value inside and outside parentheses in g(x) gives ...
... g(x) = -1(x² -6x +9) -5 +9
... g(x) = -(x -3)² +4 . . . . . vertex form
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<em>About transformations</em>
g(x) = f(x -a) causes the graph of f(x) to be shifted "a" units to the right. For a function f(x) with an axis of symmetry at x=0, it moves the axis of symmetry to x=a.
g(x) = f(x) +a causes the graph of f(x) to be shifted "a" units up.
g(x) = -f(x) causes the graph of f(x) to be reflected across the x-axis.
Here, we have all three of these transformations. First is the reflection:
... f₁(x) = -f(x) = -x²
Then we have shifting to the right 3 units. (also moves the axis of symmetry)
... f₂(x) = f₁(x-3) = -(x -3)²
Finally, we have shifting up 4 units.
... g(x) = f₂(x) +4 = -(x -3)² +4