Show that if y=x^2 + 8x -3

then y is greater to -19 for all values of x.

It comes from the topic of Completing the Square

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It is given that y=x^2 + 8x -3

y=x^2 + 8x -3

=> y=x^2 + 8x + 16 -3 - 16

=> y = (x + 4)^2 - 19

As (x + 4)^2 cannot be negative it can take a minimum value of 0, y is greater than or equal to -19.

**This proves that y is >= -19 for all values of x**

Since y = x^2 + 8x -3, we'll form the inequality:

x^2 + 8x -3 > -19

We'll move -19 to the left side:

x^2 + 8x -3 + 19 > 0

x^2 + 8x + 16 > 0

We'll verify if the quadratic equation has solutions:

x^2 + 8x + 16 = 0

x1 = [-8+(64-64)]/2

x1 = x2 = -4

**The quadratic x^2 + 8x + 16 is positive over the intervals (-`oo` ; -4) U (-4 ; `oo` ), except for the value x = -4, where the quadratic is cancelling. **

The question is asking to prove a minimum value for the equation. Special forms like radicals and squares would help in this situtation. In this particular equation, since there is no redicals, the only form is a square. In order to only leave a constant left to get the minimum value, we need to use both x^2 and 8x.

Using logic and maybe plugging in, we find that the only square form of it is (x+4)^2 which equals x^2+8x+16

However, we have -3 in this case

split -3 into 16 and -19 (16-19=-3)

we have

x^2+8x+16-19

put the form into a square

(x+4)^2-19

we know tha a sqaure could only be greater or equal to 0

(x+4)^2>=0

minus 19 on both sides

(x+4)^2 -19 >= -19

**This shows that the f(x) or y value must be bigger or equal then -19. I think you mistyped the question since the function could EQUAL -19, it could be proved that y= x^2+8x-3 , y value is always greater than or equal to -19.**

Hope This Helps

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