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)
put the form into a square
we know tha a sqaure could only be greater or equal to 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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