The diagram includes a parabola with the point (6, 1).
“The figure above shows the graph of the quadratic function f(x)=a(x-4)^2+b, where a and b are constants. Which of the following number lines represents the set of all values that satisfy the inequality f(x)=a(x-4)^2+b <= 1?”
The answer was between 2 and 6 inclusive, but not sure how to get to it? (May 2013 SAT)
Not sure if when i submitted it on the other q and a it went through so posting here incase it didn’t.
You’re looking for all the values of x for which the parabola is below the y = 1 line. Draw the y = 1 line (it’s horizontal, and goes through that point (6, 1) you’re given—that’s why that’s the point that’s given). You’ll see that the parabola is below the line between x = 2 and x = 6, inclusive.
If the figure doesn’t give enough information for you to solve just by eyeballing it, then note that the function is written to show you clearly that the parabola is translated 4 to the right. That tells you the vertex is at x = 4. Symmetry tells you that if the parabola goes through (6, 1) and has a vertex at x = 4, then it also goes through (2, 1).
Thanks, great explanation… that vertex idea totally slipped my mind