A function is defined as f(x)=ax^2+bx+d, where a, b, and d are integers.
The minimum value of f(x) is -4
Determine the x-coordinate of the minimum value of f(x) and hence find the value of d.

Question

anonymous · Answer

okay, well you know that at some x the minimum is -4, so you can find that point by taking the derivative of f(x) and setting it to 0 which is: $$2ax +b = 0$$ so $$x = -b/2a$$ then you can place that in the first equation: $$ax^2 + bx + d = a(-b/2a)^2 + b(-b/2a) + d = b^2/2a -b^2/2a + d = -4$$ thus $$d = -4$$  then you can find x by setting f(x) = -4 and then solving for x, which gives $$-4 = ax^2 +bx -4 --> 0 = ax^2+bx = x(ax + b)$$ so you can say that $$x = 0, -b/a$$

anonymous · Answer

sorry haha, math error :O. just noticed.

should be $$b^2/4a^2 - b^2/2a + d = -4$$ which would change everything. its kinda late here......sorry.

anonymous · Answer

so then $$x = -b/2a$$ and $$d = (-16a^2 +b^2(2a-1))/4a^2$$

anonymous · Answer

not $$4a^2$$ i meant $$4/a$$....I am going to bed.

anonymous · Answer

So to actually put it together nicely. $$x=-b/2a$$ and $$d = -4 - b^2/4a + b^2/2a = -4 + b^2/4a$$.