Let's begin by rewriting our function in the general expression of a quadratic function, y=ax^2+bx+c.
f(x) = x^2+3x-12⇔ f(x) = 1x^2 + 3x + (- 12)
We can see that a = 1, b = 3, and c = - 12.
Determining Maximum or Minimum Value
Before we determine the maximum or minimum recall that, if a>0, the parabola has a minimum value. Conversely, if a<0, the parabola has a maximum value.
In the given function, we have a=1, which is greater than 0. Thus, the parabola will have a minimum value.
Finding the Maximum or Minimum Value
The minimum or maximum value of a parabola is always the y-coordinate of the vertex. We can find it by first looking for the x-coordinate of the vertex, - b2a.
Unless there is a specific restriction given in the context of the problem, the domain of a quadratic function is all real numbers. In this case, there is no restriction on the value of x. Since the minimum value of the function is - 14.25, the range is all real numbers greater than or equal to - 14.25.
Domain:& {All real numbers}
Range:& {f(x)|f(x)≥ - 14.25}
