Question

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

Answer 1

Step-by-step explanation

• Method 1 (Formula)

`\begin{cases}h = - (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 = - (5)/(2(1)) \\ h = - (5)/(2)`

k-value

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

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

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

Answer

`\sf{the \: \: minimum \: \: value \: \: is \: \: - (49)/(4) \: \: at \: \: x = - (5)/(2) }`