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avatar+19 

What is the smallest possible real value of \(x^2+8x\)?

 Apr 24, 2019
 #1
avatar+2547 
+3

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.

 Apr 24, 2019
 #2
avatar+6046 
+2

\(\text{For this problem the simplest route is just completing the square}\\ y=x^2+8x = \\ x^2 + 8x + 16 - 16 = \\ (x+4)^2 - 16\\ \text{This is a minimum at }x=-4 \text{ and the minimum is }y=-16\)

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 Apr 24, 2019
 #3
avatar+2547 
+3

So since we are trying to find the value of the expression, does that mean that the Y value is the answer?

CalculatorUser  Apr 24, 2019
 #4
avatar+6046 
+1

your answer was correct, and guessing has it's place in solving problems,

but there was a concrete way to solve this one.

Rom  Apr 25, 2019
 #5
avatar+107060 
+2

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. 

Melody  Apr 25, 2019

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