Question

Find the minimum value of the function f(x)=x^2+5x-6

Answer 1

Explanation

• Method 1 (Formula)

$\begin{cases}h = - \frac{b}{2a} \\ k = \frac{4ac - {b}^{2} }{2a} \end{cases}$

The vertex of Parabola is the maximum/minimum point depending on the value of a.

• Find Vertex

h-value

$h = - \frac{5}{2(1)} \\ h = - \frac{5}{2}$

k-value

$k = \frac{4(1)( - 6) - {(5)}^{2} }{4(1)} \\ k = \frac{ - 24 - 25}{4} \\ k = \frac{ - 49}{4} \\ k = - \frac{49}{4}$

The minimum value is the value of k. Therefore the minimum value is - 49/4 at x = -5/2.

• Method 2 (Derivative)

This is Calculus method. We simply differentiate the function then substitute y' = 0.

• Differentiate Function

$f'(x) = 2 {x}^{2 - 1} + {5x}^{1 - 1} - 0 \\ f'(x) = 2x + 5$

Substitute f'(x) = 0

$0 = 2x + 5 \\ - 5 = 2x \\ - \frac{5}{2} = x$

Substitute x = -5/2 in the original equation.

$f(x) = {( - \frac{5}{2} )}^{2} + 5( - \frac{5}{2} ) - 6 \\ f(x) = \frac{25}{4} - \frac{25}{2} - 6 \\ f(x) = \frac{25}{4} - \frac{50}{4} - \frac{24}{4} \\ f(x) = \frac{25}{4} - \frac{74}{4} \\ f(x) = - \frac{49}{4}$

Answer

$\sf{the \: \: minimum \: \: value \: \: is \: \: - \frac{49}{4} \: \: at \: \: x = - \frac{5}{2} }$