To answer this question, we need to observe with attention the vertex for each function. This is the maximum value for a parabola. Then, we can determine the maximum for the function on the graph, and then we can determine algebraically the vertex for the second function.
First Case: Graphed Function
We have the graph of this function as follows:
If we identify the vertex for this function, we can see that the vertex is the point x = 4, and y = 1 or (4, 1).
Second Case: The function -x²+2x-15
We can use the vertex formula for a parabola as follows:
[tex]x_v=-\frac{b}{2a},y_v=c-\frac{b^2}{4a}[/tex]
This formula is for a parabola equation of the form:
[tex]ax^2+bx+c[/tex]
Since we have the function:
[tex]-x^2+2x-15^{}[/tex]
Then, we have that:
[tex]a=-1,b=2,c=-15[/tex]
Therefore, the vertex is:
[tex]x_v=-\frac{b}{2a}\Rightarrow x_v=-\frac{2}{2(-1)}\Rightarrow x_v=-\frac{2}{-2}\Rightarrow x_v=1[/tex]
And the y-value is:
[tex]y_v=c-\frac{b^2}{4a}\Rightarrow\begin{cases}a=-1 \\ b=2 \\ c=-15\end{cases}[/tex][tex]y_v=-15-\frac{2^2}{4(-1)}=-15-\frac{4}{-4}=-15+\frac{4}{4}=-15+1=-14[/tex]
Then, the vertex is (1, -14).
If we compare the two functions, we have:
• The vertex of function 1 is (4, 1).
,
• The vertex of function 2 is (1, -14).
In summary, therefore, the function that has a greater maximum value is function 1 [the value for the vertex is (4, 1)].
