Question

what is the minimum/ maximum value of the parabola and how was it determined? x^2+ 5x - 24 = 0

Answer 1

ANSWER

-30.25

EXPLANATION

The maximum or minimum value of a parabola is the y-coordinate of the vertex. If the parabola is given in the standard form,

$f(x)=ax^2+bx+c$

Then the x-coordinate of the vertex is,

$x_{vertex}=\frac{-b}{2a}$

And the y-coordinate of the vertex is,

$y_{vertex}=f(x_{vertex})$

In this case, a = 1, and b = 5, so the x-coordinate of the vertex is,

$x_{vertex}=\frac{-5}{2\cdot1}=-\frac{5}{2}$

And the y-coordinate is,

$y_{vertex}=\left(-\frac{5}{2}\right)^2+5\left(-\frac{5}{2}\right)-24=\frac{25}{4}-\frac{25}{2}-24=\frac{25-50-96}{4}=\frac{-121}{4}=-30.25$

Hence, the minimum/maximum value of this parabola is -30.25.