Write three different quadratic functions in vertex form that open downward

Question

Write three different quadratic functions in vertex form that open downward- One with two zeros, one with one zero, and one with no zeros.

Explain how you connected the number of zeros to the equation.

Answer 1

y=-x^2 has one zero, the origin (0,0) the x^2 term has to have a negative coefficient to open downward. the x intercept or zero is the vertex = (0,0) It's in vertex form with y=-(x-0)^2 + 0 where vertex is (0,0) which is the maximum point of the parabola. Although it has only one zero, its a zero with multiplicity 2. whenever the multiplicity is 2, the curve doesn't intersect the x axis, but just touches it. It's tangent to the x axis. You could say it really as 2 zeros, but the two zeros are identical, both the same point.

y=-(x-4)^2 + 4 has 2 zeros, x= 6 and x= 2, or (6,0) and (2,0)

Either x value makes y=0. the quadratic, parabola, opens downward due to the - in front of (x-4)^2 It's in vertex form with (4,4) as the vertex which is the maximum point of the parabola

y=-x^2 - 4 has no zeros. the curve lies everywhere below the x -axis

It never crosses the x axis. It has no real zeros. But it does have imaginary zeros. x= + and - 2i

y=-(x-0)^2 -4 is in vertex form with vertex = (0,-4) = maximum point of the parabola