What is the greatest integer value of b such that -4 is not in the range of ?
Thank you!!
Since the parabola turns upward, we re looking for the y value of a vertex that is > - 4
So....we can solve the following to find the x coordinate of this vertex
-b/ (2 * a ) = x coord of the vertex
-b / (2 * 1) = x
-b/2 = x
And we require that
x^2 + bx + 12 > - 4 substituting, we have that
(-b/2)^2 + b (-b/2) + 12 > -4
b^2 /4 -b^2/2 > -16
-b^2 / 4 > -16
-b^2 > -64 multiply by -1 and reverse the inequality sign
b^2 < 64
So
-8 < b < 8 will produce a parabola whose range > - 4
And.....the graph is still in range whenever the largest integer value of b = 7
For comparative purposes......see the graphs here for some different values of "b"
https://www.desmos.com/calculator/my77qzagdu
Note that when l b l < 8 the range is > -4
But when l b l ≥ 8, the graph is out of range
Since the parabola turns upward, we re looking for the y value of a vertex that is > - 4
So....we can solve the following to find the x coordinate of this vertex
-b/ (2 * a ) = x coord of the vertex
-b / (2 * 1) = x
-b/2 = x
And we require that
x^2 + bx + 12 > - 4 substituting, we have that
(-b/2)^2 + b (-b/2) + 12 > -4
b^2 /4 -b^2/2 > -16
-b^2 / 4 > -16
-b^2 > -64 multiply by -1 and reverse the inequality sign
b^2 < 64
So
-8 < b < 8 will produce a parabola whose range > - 4
And.....the graph is still in range whenever the largest integer value of b = 7
For comparative purposes......see the graphs here for some different values of "b"
https://www.desmos.com/calculator/my77qzagdu
Note that when l b l < 8 the range is > -4
But when l b l ≥ 8, the graph is out of range