y=c(x^2) + d
In the systems of equations above, c and d are constants . For which of the following values of c and does the system of equations have no real solutions?
A) c=-6, d=6
B) c=-5, d=4
C) c=6, d=4
D) c=6, d=5

Please help, thank you.

The system will have no real solutions when the graphs of the two equations don’t intersect. Easy one first: 2y=10 simplifies to y=5, a horizontal line.

So we need to pick values for c and d that will prevent the parabola from intersecting the y=5 line.

Quick way to nail it is to remember that for a parabola in y=ax^2+bx+c form:

  • The leading coefficient tells you whether the parabola opens up (if it’s positive) or down (if it’s negative).
  • The constant term tells you the y-intercept.

So we need either a y-intercept greater than 5 with a parabola that opens up, or a y-intercept less than 5 with a parabola that opens down.

A) No, goes through (0, 6) and opens down.
B) YES! Opens down, has a y-intercept of 4.
C) No, goes through (0, 4) and opens up.
D) No, goes through (0, 5) so obviously intersects the y=5 line.

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