Dayaan M. answered • 08/26/26
Earned A’s Twice in Precalculus | 5 Years of Tutoring Experience
In order to draw a graph from a list of conditions like this, do not try to picture the whole thing at once. Take the conditions one at a time and let each one tell you about one feature, then put the pieces together at the end.
Start with the second derivative, since it controls the shape. f''(x) > 0 means concave up, which you can picture as the curve holding water like a cup. f''(x) < 0 means concave down, like an upside down dome. So we need the curve cupping upward to the left of x = 1, flipping to a dome between 1 and 3, and then cupping upward again after 3.
Because the concavity actually changes at x = 1 and at x = 3, both of those are inflection points. That is worth marking on your sketch before you draw anything.
Now the first derivative. f'(x) = 0 means a horizontal tangent, so the curve flattens out at x = 1 and at x = 6.
Here is the interesting part, and it is the whole point of the problem. At x = 6 we have f'(6) = 0 and we are in the region where f'' > 0, so the curve is concave up there. A flat spot at the bottom of a cup is a local minimum. But at x = 1 we have f'(1) = 0 AND the concavity is changing. That combination does not give you a max or a min. It gives you a spot where the curve momentarily flattens out and then keeps going the same direction, which is sometimes called a saddle point or a horizontal inflection point. Students almost always want to draw a max or min at x = 1, and that is the trap here.
Last, no x-intercepts just means the curve is never allowed to touch the x-axis. The easiest way to satisfy that is to keep everything comfortably above it.
So putting it together, sketch it like this. Start high on the left, coming down while concave up. Flatten out momentarily at x = 1 but keep heading down through it, since it is a horizontal inflection point rather than a bottom. Between x = 1 and x = 3 keep falling but now concave down, like going over the shoulder of a hill. At x = 3 the concavity flips back to concave up while you are still falling. Keep descending until you flatten out and turn around at x = 6, which is your local minimum, then rise after that. Keep the entire curve above the x-axis the whole way.
So, our final answer is a curve that is decreasing from the left through a horizontal inflection point at x = 1, has inflection points at both x = 1 and x = 3, bottoms out at a local minimum at x = 6, rises afterward, and stays entirely above the x-axis.
