Identify the transformations for the quadratic function y= -2(x-1)^2 + 6

Answer 1

That answer was right by Kevin C. The question, "what are the transformations occurring to the graph of

y= -2(x-1)^2 + 6 " can be described by comparing the graph to a graph of y = x^2, a parabola. Here, the general equation y = a(x-h)^2 + k (where (h,k) is the vertex, and a is nonzero) still applies, but in the case of y = x^2, h and k are both 0, and a = 1. Therefore the graph has a vertex at (0,0) and it opens upwards with points on the graph at (1,1), (2,4), (-1,1) and (-2,4). To transform this graph into y = -2(x-1)^2 + 6, we need to provide a horizontal translation of +1 unit, meaning shift the graph of y = x^2 one unit to the right so that its axis of symmetry is no longer the y axis but instead is the line x = 1 (P.S. even though we are subtracting 1 from x in the equation (x-1), you still shift the graph of y=x^2 to the right; similarly if we were to add +1 to x

(i.e. x+1) in the equation, we would shift y=x^2 to the left by -1 units). The graph always moves to the opposite of the apparent operation (subtraction or addition) when we are dealing with x.

That is not true for the y value, however. The +6 provides a vertical shift of +6 units to the graph, so its vertex will be at (1, 6). The a = -2 value flips the entire graph upside down over its vertex (because of the negative sign), so instead of opening upward, the graph now opens downward. Also because of the -2 value in a = -2, the graph will be stretched by a value of -2, downward. If we were dealing with the equation

y = -(x-1)^2 + 6 (the equation, without the -2; a = -1), the equation has a vertex of (1,6). You can expect this graph to have points at (0,5) and (2,5). However since we are really concerned about the graph -2(x-1)^2 + 6 (the real equation in question; the equation with a = -2), the vertex is still at (1,6) but the same comparative points are at (0,4) and (2,4).

In conclusion, the transformations that take place to the graph of y = x^2 to become y = -2(x-1)^2 + 6 are a horizontal translation of +1 units to the right, a vertical translation of +6 units upwards, and because of a = -2, the flipping of the graph over its vertex (1,6). It shifts from opening upward to downward (because a = -2 is negative), and a stretching of the graph downward by a factor of -2 since a = -2.

Answer 2

The general vertex form a quadratic equation is y=a(x-h)^2+k, where (h,k) is the vertex, "a" indicates a stretch or shrink, a negative value of "a" indicates reflection about the x-axis, and the axis of symmetry is x=h. Horizontal translations are associated with h, using the opposite sign. Vertical translations are associated with k, using the same sign.

Here, we have: a=-2, h=1, k=6.

So the transformations, in no particular order, are: Stretch by factor of 2, reflect about x-axis, right 1, up 6.

Answer 3

Hello,

I have worked out this problem in this video. When time permits, please take a moment to review it and let me know if you have any additional questions. I hope this helps.

Take care,

Dr. Christal-Joy Turner

Answer 4

See the videa.