This assignment starts by investigating y = x^2 + bx + 1 for b = -3, -2, -1, 0, 1, 2, 3
See the graphs below:
The vertex of each parabola listed can be found by using the formula -b/2a. The calculations for -b/2a gives the x coordinate. Plug the x coordinate back into the equation and solve for y. The vertex of the above equations are as follows:
purple: y=x^2-3x+1 (3/2,-5/4)
red: y=x^2-2x+1 (1,0)
blue: y=x^2-1x+1 (1/2,3/4)
green: y=x^2-0x+1 (0,1)
light blue: y=x^2+x+1 (-1/2, 3/4)
yellow: y=x^2+2x+1 (-1,0)
grey: y=x^2+3x+1 (-3/2, -5/4)
See the vertices graphed on the parabolas below:
The locus of the vertices of the set of parabolas graphed above is the parabola:
The parabola y=-x^2+1 is going in the opposite direction. Hence, the negative. In order to find the equation algebraicly, plug the vertex and a point on the parabola into the vertex formula
y = a(x-h)^2+k, (h.k) is the vertex
y = a(x-0)^2+1
y = ax^2+1
The I chose was (-1,0) in order to find a
0 =a(-1)^2 +1
0=a +1
-1 = a
Therefore, y=-x^2 + 1 only makes sense.