a) In the parabola y=x^2 the vertex is at the origin (0,0). Rewrite the other equation (y+49)=(x-2)^2. This time the vertex is at the point (2,-49), which means that in the transformation, the vertex moves 49 units down from the origin and 2 units to the right. The axis of symmetry, which was the y axis is now x=2.
b) y=3x+a where a is a constant represents all incoming signals. To intersect the parabola twice, substitute for y in the parabola equation: 3x+a=x^2-4x+4-49. So x^2-7x-45-a=0, and x=(7±√(49+180+4a))/2. Look at the expression under the square root: 229+4a. This expression must be positive: 229+4a≥0. Also for two intersections 229+4a>0 so that we have two roots. Therefore 4a>-229, a>-229/4, a>-57.25. Example: a=-1, x=(7±15)/2=11 and -4, y=32 and -13. The intersection points for y=3x-1 are (11,32) and (-4,-13) (see graph below, showing part of the parabola in red and the signal in blue).