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To determine whether the quadratic function [tex]\( f(x) = 3x^2 - 24x - 5 \)[/tex] has a maximum or minimum value, we can analyze the properties of the quadratic function.
1. We observe that the coefficient of [tex]\( x^2 \)[/tex] in the quadratic function is positive ([tex]\( 3 > 0 \)[/tex]). For quadratic functions of the form [tex]\( ax^2 + bx + c \)[/tex], if [tex]\( a > 0 \)[/tex], the parabola opens upwards, indicating that the function has a minimum value (since the vertex of the parabola represents the lowest point).
2. Next, we find the coordinates of the vertex of the parabola. The x-coordinate of the vertex of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = 3 \)[/tex] and [tex]\( b = -24 \)[/tex]. Substituting these values into the formula gives:
[tex]\[ x = -\frac{-24}{2 \cdot 3} = \frac{24}{6} = 4 \][/tex]
3. To find the y-coordinate of the vertex, we substitute [tex]\( x = 4 \)[/tex] back into the original function [tex]\( f(x) \)[/tex]:
[tex]\[ f(4) = 3(4)^2 - 24(4) - 5 \][/tex]
Calculating further:
[tex]\[ f(4) = 3 \cdot 16 - 24 \cdot 4 - 5 = 48 - 96 - 5 = -53 \][/tex]
Hence, the vertex of the quadratic function [tex]\( f(x) = 3x^2 - 24x - 5 \)[/tex] is at [tex]\( (4, -53) \)[/tex]. Since the parabola opens upwards, this vertex represents the minimum value of the function.
Thus, the function [tex]\( f(x) \)[/tex] has a minimum value of [tex]\(-53\)[/tex].
The function has a [tex]\(\text{minimum}\)[/tex] value of [tex]\(\text{-53}\)[/tex].
1. We observe that the coefficient of [tex]\( x^2 \)[/tex] in the quadratic function is positive ([tex]\( 3 > 0 \)[/tex]). For quadratic functions of the form [tex]\( ax^2 + bx + c \)[/tex], if [tex]\( a > 0 \)[/tex], the parabola opens upwards, indicating that the function has a minimum value (since the vertex of the parabola represents the lowest point).
2. Next, we find the coordinates of the vertex of the parabola. The x-coordinate of the vertex of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = 3 \)[/tex] and [tex]\( b = -24 \)[/tex]. Substituting these values into the formula gives:
[tex]\[ x = -\frac{-24}{2 \cdot 3} = \frac{24}{6} = 4 \][/tex]
3. To find the y-coordinate of the vertex, we substitute [tex]\( x = 4 \)[/tex] back into the original function [tex]\( f(x) \)[/tex]:
[tex]\[ f(4) = 3(4)^2 - 24(4) - 5 \][/tex]
Calculating further:
[tex]\[ f(4) = 3 \cdot 16 - 24 \cdot 4 - 5 = 48 - 96 - 5 = -53 \][/tex]
Hence, the vertex of the quadratic function [tex]\( f(x) = 3x^2 - 24x - 5 \)[/tex] is at [tex]\( (4, -53) \)[/tex]. Since the parabola opens upwards, this vertex represents the minimum value of the function.
Thus, the function [tex]\( f(x) \)[/tex] has a minimum value of [tex]\(-53\)[/tex].
The function has a [tex]\(\text{minimum}\)[/tex] value of [tex]\(\text{-53}\)[/tex].
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