Let's analyze the given quadratic function [tex]\( r(t) = -3.22t^2 + 40.63t + 11.17 \)[/tex] in detail to select the correct graph.
1. Identify the general shape of the graph:
- The quadratic function [tex]\( r(t) = at^2 + bt + c \)[/tex] is a parabola.
- Since the coefficient of [tex]\( t^2 \)[/tex] (which is [tex]\( a \)[/tex]) is negative ([tex]\( a = -3.22 \)[/tex]), the parabola opens downwards.
2. Find the vertex of the parabola:
- The vertex formula for a quadratic function [tex]\( y = ax^2 + bx + c \)[/tex] is given by [tex]\( t = -\frac{b}{2a} \)[/tex].
- Here, [tex]\( a = -3.22 \)[/tex], [tex]\( b = 40.63 \)[/tex], and [tex]\( c = 11.17 \)[/tex].
- Compute the vertex:
[tex]\[
t = -\frac{b}{2a} = -\frac{40.63}{2 \times -3.22} = 6.309006211180124
\][/tex]
- The vertex form is when [tex]\( t = 6.309 \)[/tex].
3. Calculate the revenue at the vertex:
- Substitute [tex]\( t = 6.309 \)[/tex] back into the quadratic function to find the revenue [tex]\( r \)[/tex] at this time:
[tex]\[
r(6.309) = -3.22(6.309)^2 + 40.63(6.309) + 11.17
\][/tex]
[tex]\[
r(6.309) = 139.33746118012422
\][/tex]
- We know the exact vertex revenue is approximately [tex]\( 139.337 \)[/tex].
4. Behavior of the graph:
- The vertex [tex]\( (6.309, 139.337) \)[/tex] represents the maximum revenue point.
- The revenue starts at [tex]\( 11.17 \)[/tex] and increases to [tex]\( 139.337 \)[/tex] at [tex]\( t = 6.309 \)[/tex], then decreases thereafter.
To conclude, the graph associated with this model should:
- Be a downward-opening parabola.
- Have its peak (vertex) at approximately [tex]\( (6.31, 139.34) \)[/tex].
- Pass through the point where [tex]\( t = 0 \)[/tex] which gives [tex]\( r(0) = 11.17 \)[/tex].
This description should help in identifying the correct graph among the options provided. Be sure to look for a graph that showcases a peak value at approximately 6.31 on the t-axis and has a corresponding peak revenue near 139.34 on the r-axis.
