A function is defined by f(x) = x^2 + 4x – 5. What is the minimum valu

The function defined in the question is a quadratic function with a positive co-efficient for \(x^2\). There are two methods which can be adopted to solve this question.
Let’s discuss both of them in detail.

For a quadratic function of the form a\(x^2\) + bx + c where a>0, the minimum value of the function is given by \(\frac{4ac – b^2}{4a}\).

Comparing the function given in the question with the standard form, we have,

a = 1, b = 4 and c = -5.

Substituting these values in the expression above, we have,

Minimum value of f(x) = \(\frac{4 * 1 * (-5) – (4)^2}{4*1}\) = \(\frac{- 20 – 16}{4}\) = -9.

For a quadratic function of the form a\(x^2\) + bx + c where a>0, the minimum value of the function occurs at x = \(\frac{-b}{2a}\).

This means, for the given function, the minimum value of the function occurs at x = \(\frac{-4}{2}\) = -2.

Substituting this value in the function, we have f(-2) = 4 – 8 – 5 = -9.
Therefore, the minimum value of the given function is -9. The correct answer option is E.

Hope this helps!