+2863 Definitely not positive integers... because of the +8x
Negatives can't be too small because of the \(x^2\) will turn it into a positive integer.
Because I'm am not the best at math, I will just use a guess and check to solve.
1. We are trying to find the peak at which the value will be on the verge of becoming bigger.
Substituting -3 for x we will get : -15
Substituting -4 for x we will get : -16
Substituting -5 for x we will get : -15
So the answer SHOULD be -16? I'm not sure someone check this.
+6251 \(\text{For this problem the simplest route is just completing the square}\\ y=x^2+8x = \\ x^2 + 8x + 16 - 16 = \\ (x+4)^2 - 16\\ \text{This is a minimum at }x=-4 \text{ and the minimum is }y=-16\)
.+2863 So since we are trying to find the value of the expression, does that mean that the Y value is the answer?
+6251 your answer was correct, and guessing has it's place in solving problems,
but there was a concrete way to solve this one.
+118677 Hi CalculatorUser,
A big thanks to you and Rom.
To answer your question... Yes it does.
y= x^2+8x is a concave up parabola. (so it will have a minimum value)
y=x(x+8)
So the 2 roots are at x=0 and x=-8
The axis of symmetry is half way between them x=(0+-8)/2 = -4
The minimum will be the y value when x=-4
y=-4(-4+8) = -4*4 = -16
sketch what I have done so you can see what I am talking about.
