If parallelogram ABCD was reflected over the y-axis, reflected over the x-axis, and rotated 180°, where would point A′ lie?
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Answer:
<h3>Question 1</h3><u>Function given:</u>
- y = x² + 4x - 4
<u>In the vertex form:</u>
- y = x² + 4x + 4 - 8 = (x + 2)² - 8
Axis of symmetry is the vertical line passing through the vertex.
<u>The vertex is</u>
- (-2,-8)
<u>Axis of symmetry is </u>
- x = -2
<u>y- intercept is </u>
- x = 0 ⇒ y = -4, so the point (0, -4)
<u>Two other points are</u>
- x = 1 ⇒ y = (1 + 2)² - 8 = 9 - 8 = 1, the point is (1, 1)
- x = -1 ⇒ y = (-1 + 2)² - 8 = 1 - 8 = -7, the point is (-1, -7)
<em>See the graph with all the points plotted</em>
<h3>Question 2</h3><u>Function given:</u>
- y + 2x² - 4x - 6 = 0
<u>In standard form:</u>
- y = -2x² + 4x + 6
<u>In vertex form:</u>
- y = - 2x² + 4x - 2 + 8 = -2(x - 1)² + 8
<u>The vertex:</u>
- (1, 8)
<u>Axis of symmetry is </u>
- x = 1
<u>y - intercept </u>
- x = 0 ⇒ y = -2(-1)² + 8 = -2 + 8 = 6, the point is (0, 6)
<u>Two other pints:</u>
- x = -1 ⇒ y = -2(-1 - 1)² + 8 = -8 + 8 = 0, the point is (-1, 0)
- x = 2 ⇒ y = -2(2 - 1)² + 8 = -2 + 8 = 6, the point is (2, 6)
<em>The graph is attached with the points plotted.</em>
