The quadratic equation is f(x) = (x + 12)² + 20, and the numerical values of h and k are h = -12 and k = 20
How to determine the values of h and k?
The function is given as:
f(x) = x² + bx + 164
Start by differentiating the function
f'(x) = 2x + b
Set to 0
2x + b = 0
Solve for x
x = -b/2
Substitute x = -b/2 in f(x) = x² + bx + 164
f(-b/2) = (-b/2)² + b(-b/2) + 164
This gives
f(-b/2) = b²/4 - b²/2 + 164
Evaluate
f(-b/2) = - b²/4 + 164
The minimum of the function is 20.
So, we have:
- b²/4 + 164 = 20
Subtract 164 from both sides
- b²/4 = - 144
Multiply both sides by -4
b² = 576
Take the square root of both sides
b = 24
Substitute b = 24 in f(x) = x² + bx + 164
f(x) = x² + 24x + 164
Rewrite as:
f(x) = (x² + 24x) + 164
This gives
f(x) = (x² + 24x + (24/2)² - (24/2)²) + 164
Rewrite as:
f(x) = (x² + 24x + 12²) - 12² + 164
f(x) = (x² + 24x + 12²) + 20
Express as a perfect square expression
f(x) = (x + 12)² + 20
The function is represented as:
f(x) = (x - h)² + k
By comparison, we have:
h = -12 and k = 20
Hence, the numerical values of h and k are h = -12 and k = 20, respectively
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