I just need help getting started on this problem. I don't even know how to begin!
The graph is centered over the line x=-3. This line acts like a mirror so that the two halves of the inverted U-shaped graph are reflected in it. The graph has to be shifted 3 units to the right so that the line x=-3 becomes the line x=0, which is the y (f(x)) axis. The equation of the graph changes to f(x)=-0.5x^2+8. When x=0 f(0)=8, and this value on the vertical axis is the highest point of the curve (vertex).
To picture f(x+k) you need a picture, a moving picture. Think of the horizontal axis, the x axis, and the line f(x)=8, which is a line parallel to the x axis a distance of 8 units above it. The two lines resemble a track, like a rail track. The curve is, as we've established, an inverted U shape where the arms of the U are moving further apart the further away the curve is from its vertex. The curve has a constant shape and remains in contact with the line f(x)=8. We start with the curve right in the middle so that the vertical axis bisects the U curve. This is when k=0.
Now, we're going to slide the curve so that its vertex runs along the track f(x)=8 left and right. The curve cuts the axis at two points separated by 8 units, the gap between the zeroes of the function. When the curve is centrally positioned, the points on the x axis are -4 and 4 and k=0. Slide the curve to the left (negative side) and this is equivalent to positive values of k; slide it to the right and we're into negative values of k. If we move the y (f(x)) axis with the curve it becomes the movable axis of symmetry, the mirror I mentioned earlier. The points where the curve cuts the x axis remain 8 units apart. When the curve moves leftward one unit, the value of k increase by 1, and rightward one unit when k decreases by 1.
The vertical line x=2 is unmovable, but as the curve slides from left to right this line touches or cuts through the curve. Consider only the part of the curve lying on or above the x axis. When does it touch the vertical line? Move the curve so that the right part of the curve just touches the line. This is the zero with the higher value. So it must be when x=2 is a root, i.e., -0.5(2+k)^2+8=0. 0.5(2+k)^2=8, so (2+k)^2=16, 2+k=+4, and k=4-2 or -4-2, which is 2 or -6.
If we slide the curve past the line x=2, till the left part of the curve touches the line on the right side, the point where it touches is the other root, 8 lower than the the right-hand root. So if 2 was the right root then 2-8=-6 is the left root, and if 2 was the left root then 10 is the right root. So we have the functions f(x)=-0.5(x+2)^2+8 and f(x)=-0.5(x-6)^2+8, where the values of k have been substituted. When the curve is on the left of x=2, the roots of f(x+2) are -6 and 2 and when on the right the roots of f(x-6) are 2 and 10.
In between these values of k, 2 and -6, the curve touches or is above the x axis, so this the range for k: -6 Read More: ...