Question

Let ​ f(x)=x2−8x+19 What is the minimum value of the function?​

Answer 1

By completing the square: x²-8x+19 = (x-4)²-16+19 = (x-4)²+3 so minimum point = (4,3)

By calculus: f'(x) = 2x-8. (minimum point occurs when f'(x)=0 therefore)
2x-8 = 0 means x=4 and the corresponding y value is 3.

The minimum value of this function (y) is 3 and this occurs when x = 4.

Answer 2

In order to find the minimum value or the 'vertex' in this case, we need to take the first derivative and second derivative.
The first derivative informs us of stationary points. Because this is a quadratic, there should be a point where the curve comes back up.
The second derivative determines the concavity of the stationary point. If the second derivative is less than zero, there lies a maximum turning point (ie a maximum value), and if the second derivative is greater than zero, there lies a minimum turning point (ie a minimum value).

$f(x) = x^{2} - 8x + 19$
$f'(x) = 2x - 8$

Now, we let the first derivative equal to zero, because that's when we have the tip of the slope or the horizontal tangent of the parabola.
$f'(x) = 0; 2x - 8 = 0; x - 4 = 0$

So, we know there is a horizontal tangent at x = 4.
Let's make sure this is a minimum value.

$f''(x) = 2$
Since the second derivative is greater than zero, there lies a minimum value at x = 4.

Hence, the minimum value of the function is x = 4