Find the value of a such that h(x) = ax^2 + 2ax + 4 will have a maximum value of 7?
How do I go about doing this? I've been stuck for almost an hour with this problem.
The standard way of determining a maximum value for a quadratic is to complete the square: $\displaystyle ax^2+ 2ax+ 4= a(x^2+ 2x)+ 4$$\displaystyle = a(x^2+ 2x+ 1- 1)+ 4= a(x^2+ 2x+ 1)- a+ 4$$\displaystyle = a(x^2+ 2x+1)- a+4= a(x+1)^2- a+4$. Since a square if never negative, as long as a itself is negative, that is -a+ 4 minus something. That is, its maximum value occurs at x= -1 and is equal to -a+ 4.
For $\displaystyle a\neq 0\,,\,\,ax^2+2ax+4$ is a parabola. A parabola has a maximum value if and only if it is a concave-down parabola, this is: if and only if $\displaystyle a<0$ (and this already is a limitation to a).
In this case, the maximal point is at the parabola's vertex, which in this case is given by $\displaystyle \left(-2\,,-\frac{4a^2-16a}{4a}\right)$ , with the y-entry being the maximal value. Well, now solve your problem.
Tonio