Show that if y=x^2 + 8x -3then y is greater to -19 for all values of x.It comes from the topic of Completing the Square

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