   SEARCH HOME Math Central Quandaries & Queries  Question from James, a student: graph y=(2x^2-3x)e^ax Firstly, it's not clear whether you want this for one value of a or for a set of different values. A good representative set should include positive, negative, and zero values of a. It should also give values of e^ax that will neither explode or disappear into the x-axis within too much of the range where the polynomial part, p(x) = (2x2 - 3x), is interesting. That factors as (2x - 3)x so the range should include the roots at 0 and 2/3. One decent choice would be -2 < x < 2, a = -1,0,1.

Now, sketch the parabola y = p(x) = (2x - 3)x which is upward curving and crosses the x axis at 0 and 2/3. At -2 it takes the value 14; at 2, 2.

Now pick one value for a; the easy one is 0. e0x = 1 for all x, so you have already sketched the a=0 case. Try again with a=1. Find a few values of ex in the interval [-2,2]; either use a calculator or estimate e ~ 3 for a rough sketch. Find or estimate e-2 p(-2), e-1 p(-1), e0 p(0)=0, e1/2 p(1/2), e2/3p(2/3=0, e1p(1), and e2p(2), and joint them up with a smooth curve. If you are using calculus techniques differentiate (2x2-3x)ex and set it equal to 0 to find the local maximum and local minimum. Note that exponential functions "overpower" polynomials so ex p(x) -> 0 as x->- infinity [horizontal asymptote]. Sketch, trying for a clean smooth freehand curve. Repeat for a=-1.

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