Determine whether the given quadratic function has a minimum value or maximum value. Then find the coordinates of the minimum or maximum point.

[tex]f(x) = 3x^2 + 3x - 9[/tex]

A. Maximum; [tex]\left(-\frac{39}{4}, -\frac{1}{2}\right)[/tex]
B. Maximum; [tex]\left(-\frac{1}{2}, -\frac{39}{4}\right)[/tex]
C. Minimum; [tex]\left(-\frac{39}{4}, -\frac{1}{2}\right)[/tex]
D. Minimum; [tex]\left(-\frac{1}{2}, -\frac{39}{4}\right)[/tex]



Answer :

To determine whether the given quadratic function [tex]\( f(x) = 3x^2 + 3x - 9 \)[/tex] has a minimum value or maximum value and to find the coordinates of this point, we need to follow these steps:

1. Identify the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] from the standard quadratic form [tex]\( ax^2 + bx + c \)[/tex]. Here, we have:
[tex]\[ a = 3, \quad b = 3, \quad c = -9 \][/tex]

2. Determine if the quadratic function has a minimum or maximum value:
- If [tex]\( a > 0 \)[/tex], the parabola opens upwards, indicating a minimum value.
- If [tex]\( a < 0 \)[/tex], the parabola opens downwards, indicating a maximum value.

Since [tex]\( a = 3 > 0 \)[/tex], the parabola opens upwards, and thus the function has a minimum value.

3. Find the x-coordinate of the vertex (the point where the minimum or maximum occurs):
The formula for the x-coordinate of the vertex is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]

Plugging in the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ x = -\frac{3}{2 \cdot 3} = -\frac{3}{6} = -\frac{1}{2} \][/tex]

4. Find the y-coordinate of the vertex by substituting the x-coordinate back into the function [tex]\( f(x) \)[/tex]:
[tex]\[ y = f\left(-\frac{1}{2}\right) \][/tex]

Substitute [tex]\( x = -\frac{1}{2} \)[/tex] into the equation [tex]\( f(x) = 3x^2 + 3x - 9 \)[/tex]:
[tex]\[ f\left(-\frac{1}{2}\right) = 3\left(-\frac{1}{2}\right)^2 + 3\left(-\frac{1}{2}\right) - 9 \][/tex]

Calculate each term separately:
[tex]\[ 3\left(-\frac{1}{2}\right)^2 = 3 \cdot \frac{1}{4} = \frac{3}{4} \][/tex]

[tex]\[ 3\left(-\frac{1}{2}\right) = -\frac{3}{2} \][/tex]

Combine these with the constant term (-9):
[tex]\[ y = \frac{3}{4} - \frac{3}{2} - 9 \][/tex]

Convert [tex]\(-\frac{3}{2}\)[/tex] to quarters to add easily:
[tex]\[ -\frac{3}{2} = -\frac{6}{4} \][/tex]

Combine all terms:
[tex]\[ y = \frac{3}{4} - \frac{6}{4} - 9 = -\frac{3}{4} - 9 = -\frac{3}{4} - \frac{36}{4} = -\frac{39}{4} \][/tex]

Therefore, the function has a minimum value. The coordinates of the minimum point are:
[tex]\[ \left( -\frac{1}{2}, -\frac{39}{4} \right) \][/tex]

Thus, the correct answer is:
D. minimum; [tex]\(\left( -\frac{1}{2}, -\frac{39}{4} \right)\)[/tex]