To determine which quadratic function has the greater maximum value, let's analyze the given function:
[tex]\[ f(x) = -x^2 - 3x + 1 \][/tex]
For a quadratic function in the form [tex]\( ax^2 + bx + c \)[/tex], the maximum or minimum value occurs at the vertex. For a downward-opening parabola (where [tex]\( a < 0 \)[/tex]), the vertex represents the maximum value. The x-coordinate of the vertex for a quadratic function is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For the function [tex]\( f(x) = -x^2 - 3x + 1 \)[/tex], the coefficients are:
[tex]\[ a = -1, \quad b = -3, \quad c = 1 \][/tex]
Plugging in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] into the vertex formula:
[tex]\[ x = -\frac{-3}{2 \times -1} = -\frac{3}{-2} = \frac{3}{2} \][/tex]
So, the x-coordinate of the vertex is [tex]\( -\frac{3}{2} \)[/tex].
Next, we substitute [tex]\( x = -\frac{3}{2} \)[/tex] back into the function to find the maximum value:
[tex]\[ f\left(-\frac{3}{2}\right) = -\left(-\frac{3}{2}\right)^2 - 3\left(-\frac{3}{2}\right) + 1 \][/tex]
[tex]\[ f\left(-\frac{3}{2}\right) = -\frac{9}{4} + \frac{9}{2} + 1 \][/tex]
[tex]\[ f\left(-\frac{3}{2}\right) = -\frac{9}{4} + \frac{18}{4} + \frac{4}{4} \][/tex]
[tex]\[ f\left(-\frac{3}{2}\right) = -\frac{9}{4} + \frac{18}{4} + \frac{4}{4} \][/tex]
[tex]\[ f\left(-\frac{3}{2}\right) = \frac{13}{4} \][/tex]
Therefore, the maximum value of the function [tex]\( f(x) = -x^2 - 3x + 1 \)[/tex] is [tex]\( \frac{13}{4} \)[/tex].
In the absence of another function to compare it to, we'd assume this as the greater maximum value by default.
The quadratic function [tex]\( f(x) \)[/tex] has the greater maximum value.
