A) Express 10x-x^2-27 in form -(m-x)^2 +n
B) Hence show that 10x-x^2-27 is always negative
Thankyou
let's do it by completing the square
10x-x^2-27
= -x^2 + 10x - 27
= -(x^2 - 10x ..... ) - 27
= -(x^2 - 10x + 25 - 25) - 27
= -( (x-5)^2 - 25) - 27
= -(x-5)^2 + 25 - 27
= -(x-5)^2 - 2 -----> the form you wanted
argument:
(x-5)^2 is always positive
so
-(x-5)^2 is always negative
adding another -2 to a number already negative keeps it negative
To express the quadratic expression 10x - x^2 - 27 in the form -(m - x)^2 + n, we need to complete the square. Here are the steps to do it:
A) Completing the square:
1. Rewrite the quadratic expression without the linear term (10x - x^2 - 27) as x^2 - 10x + 27.
2. To complete the square, take half of the coefficient of x (-10/2 = -5) and square it: (-5)^2 = 25.
3. Add the terms in step 2 as well as subtract them, but maintain the overall equation balance:
- x^2 - 10x + 27
= -(x^2 - 10x + 25 - 25) + 27
= -(x - 5)^2 - 25 + 27
= -(x - 5)^2 + 2.
Therefore, the expression 10x - x^2 - 27 can be expressed in the form -(m - x)^2 + n as -(x - 5)^2 + 2.
B) To show that 10x - x^2 - 27 is always negative, we can analyze the expression -(x - 5)^2 + 2.
Since the expression is in the form -(m - x)^2 + n, we can determine its maximum value by recognizing that the squared term (x - 5)^2 is always non-negative or zero. Thus, the maximum value occurs when (x - 5)^2 = 0, which means x = 5.
Substituting x = 5 into the expression -(x - 5)^2 + 2,
= -(5 - 5)^2 + 2
= -0^2 + 2
= 2.
From this analysis, we can see that the expression -(x - 5)^2 + 2 is never positive since its maximum value is 2. Therefore, 10x - x^2 - 27 is always negative.