# Plot an apparently flat function

I would like to plot

     E^(x^2 + 7 x - 30) - 1


near 0

 Plot[E^(x^2 + 7 x - 30) - 1, {x, -5, 0}]


works but I could not see anay thing near the minimum -- 7/2 --- because the rounding effect make the function flat. How could I obtain a nice graph. Of course I have tried to set PlotRange without any success.

• The E^(-30) is a huge problem since it's such a small number. Why not plot the following version of the function instead? It is merely a shift and then a stretch, but you can always add those old numbers in: Plot[Evaluate[(f[x] + 1)/Exp[-30]], {x, -5, 0}, PlotRange -> {0, 0.0002}]. – march Sep 30 '16 at 4:57
• points = Table[{x, N[E^(x^2+7x-30)-1, 64]}, {x,-5,0,1/10}]; ListPlot[points, PlotRange -> {-1 - 11/10*10^-13, -1 + 10^-13}, Ticks -> {{-5,-4,-3,-2,-1}, {N[-1-10^-13,20], -1, N[-1+10^-13,20]}}, AxesOrigin -> {0,-1-11/10*10^-13}, Joined->True] and if you study that carefully then you might be able to use this to zoom in even further to try to see your minimum – Bill Sep 30 '16 at 5:25
• After reading all thos solutions I was doubtfull since no one show that there is trully a minimum then by mistake i fell on this very weird result: thisPlot[Evaluate[E^(x^2 + 7*x - 30)/Exp[-30]], {x, -4, -3}, PlotRange -> {0, 0.0000005}] gives nothing but Plot[Evaluate[E^(x^2 + 7*x - 30)/Exp[-30]], {x, -4, -3}, PlotRange -> {, 0.0000005}] is great --- of course Ma complains but it dos the job. WHY ? – cyrille.piatecki Sep 30 '16 at 15:03

LogPlot[E^(x^2 + 7 x - 30), {x, -5, 0},
AxesOrigin -> {-5, 10^-20},
Frame -> True, FrameLabel -> {"x", "f[x]+1"}]


beware this doesn't work well due to numerical precision issues:

 f[x_] = E^(x^2 + 7 x - 30) - 1
LogPlot[ f[x] + 1, ... ]


I'm not sure if I'm making a fool of my self but

E^(x^2 + 7 x - 30) - 1 == -1 + E^(-30 + 7*x + x^2)


True

Plot[-1 + E^(-30 + 7*x + x^2), {x, -10, 2.8}]


And take Alpha into consideration:

You are trying to find the minimum of

f[x_] := E^(x^2 + 7 x - 30) - 1


By adding one, this has a min at the same location as

f2[x_] := E^(x^2 + 7 x - 30)


By taking the log of both sides, this has a min at the same location as

f3[x_] := x^2 + 7 x - 30


This can be handled by

FindMinimum[f3[x], x]

x -> -3.5


which shows that the min of your function occurs at -3.5. Evaluating f[-3.5] gives -1, which is indeed the minimum value that f can attain.