To find the range of the quadratic function [tex]\( f(x) = 2x^2 + 4x \)[/tex], we need to determine the minimum value of the function and understand its behavior at [tex]\( \pm \infty \)[/tex].
1. Identify the coefficients: The given function is of the form [tex]\( f(x) = ax^2 + bx + c \)[/tex]. Here, [tex]\( a = 2 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = 0 \)[/tex].
2. Find the vertex: For a quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex], the vertex occurs at [tex]\( x = -\frac{b}{2a} \)[/tex].
[tex]\[
x = -\frac{4}{2(2)} = -\frac{4}{4} = -1
\][/tex]
3. Evaluate the function at the vertex: Substitute [tex]\( x = -1 \)[/tex] back into the function to find the [tex]\( y \)[/tex]-value (the minimum value of the function).
[tex]\[
f(-1) = 2(-1)^2 + 4(-1) = 2(1) - 4 = 2 - 4 = -2
\][/tex]
4. Determine the direction of the parabola: Since [tex]\( a = 2 \)[/tex] (which is positive), the parabola opens upwards. This means the function has a minimum value at the vertex and increases without bound as [tex]\( x \)[/tex] moves away from the vertex in both directions.
5. State the range: Because the minimum value of the function is [tex]\( -2 \)[/tex] and it opens upwards indefinitely, the range of [tex]\( f(x) \)[/tex] is all real numbers greater than or equal to [tex]\(-2\)[/tex].
Using interval notation, the range is:
[tex]\[
[-2, \infty)
\][/tex]
