Hi there, I'm having difficulties with this particular question

The function h: R- > R, h(t)=(t^2+at+10) e ^ - t, where a is a real constant, has a derivative h'(t) = (-t^2 + (2-a)t + (a-10))e^ -t.

Find the values of a such that
I) the graph of y=h(t) has exactly on stationery point.
ii) h'(t) <0 for all t is an element of R.

This is confusing as I don't know how to go about this question.

They give you the derivative in the form of a quadratic. The thing to do is use the

discriminant, set it to 0 and solve for a. That will be the points where there is one root of

multiplicity 2 and, therefore, one stationary point.

The discriminant is b^{2}-4ac

We have from h'(t): a=-1, \;\ b=2-a, \;\ c=a-10

So, we have:

(2-a)^{2}-4(-1)(a-10)=0

Solve for a.

If the discriminant is <0, then it has no real roots.