Answer:
C: Function 2 has the larger maximum at (2, 5).
Step-by-step explanation:
We are given two functions, with Function 1 being described by the graph and Function 2 given by the function:
[tex]f(x)=-x^2+4x+1[/tex]
Notice that the leading coefficient of Function 2 is negative. The graph of Function 1 is curving downwards. So, both functions will have maximum values.
Recall that for a parobala, the maximum (or minimum) values is the y-value of the vertex point. So, let's find the vertex for each function.
For Function 1, we can see that the vertex is at (4,1). Thus, its maximum value is y = 1.
For Function 2, we will need to work out the vertex. Recall that the vertex is given by:
[tex]\displaystyle \left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)[/tex]
Function 2 is defined by:
[tex]f(x)=-x^2+4x+1[/tex]
Therefore, a = -1, b = 4, and c = 1.
Find the x-coorindate of the vertex:
[tex]\displaystyle x=-\frac{(4)}{2(-1)}=2[/tex]
Substitute this back into the function to find the y-coordinate.
[tex]f(2)=-(2)^2+4(2)+1=5[/tex]
So, the vertex of Function 2 is (2,5). Therefore, the maximum value of Function 2 is y = 5.
Since 5 is greater than 1, the maximum value of Function 2 is greater.
The answer is choice C.
