Given f(x) = 11x^2 + 13x - 28, use the discriminant to determine whether or not f(x) ever equals -100 for any values of x.
The discriminant = b^2 - 4ac.... so we have
13^2 - 4(11)(-28) = > 0 so this tells us we have real roots
But...maybe I'm missing something.....the discriminant doesn't tell us anything about whether or not f(x) ever = -100....it just tells us whether or not we may have "real" zeroes
We'd need to find the vertex for this
The x coordinate for the vertex is given by -b/2a = -13/2(11) = -13/22
And putting this back into the function.....we have the y value at the vertex = about -31.84
And since this parabola turns "up," (-13/22, -31.84) is the lowest point on the graph.....so f(x) never = -100....
Ah...that's very clever, Melody.....that thought didn't strike me...!!!