[tex] \begin{cases}h = - \frac{b}{2a} \\ k = \frac{4ac - {b}^{2} }{2a} \end{cases}[/tex]
The vertex of Parabola is the maximum/minimum point depending on the value of a.
h-value
[tex]h = - \frac{5}{2(1)} \\ h = - \frac{5}{2} [/tex]
k-value
[tex]k = \frac{4(1)( - 6) - {(5)}^{2} }{4(1)} \\ k = \frac{ - 24 - 25}{4} \\ k = \frac{ - 49}{4} \\ k = - \frac{49}{4} [/tex]
The minimum value is the value of k. Therefore the minimum value is - 49/4 at x = -5/2.
This is Calculus method. We simply differentiate the function then substitute y' = 0.
[tex]f'(x) = 2 {x}^{2 - 1} + {5x}^{1 - 1} - 0 \\ f'(x) = 2x + 5[/tex]
Substitute f'(x) = 0
[tex]0 = 2x + 5 \\ - 5 = 2x \\ - \frac{5}{2} = x[/tex]
Substitute x = -5/2 in the original equation.
[tex]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} [/tex]
[tex] \sf{the \: \: minimum \: \: value \: \: is \: \: - \frac{49}{4} \: \: at \: \: x = - \frac{5}{2} }[/tex]
