More on Quadratic Functions

Assignment Three

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.


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