Given:
The function is:
[tex]f(x)=-(x+5)(x+1)[/tex]
To find:
The range of the function.
Solution:
We have,
[tex]f(x)=-(x+5)(x+1)[/tex]
It can be written as:
[tex]f(x)=-(x^2+x+5x+5)[/tex]
[tex]f(x)=-(x^2+6x+5)[/tex]
Add and subtract square of half of coefficient of x, i.e., [tex]\left(\dfrac{6}{2}\right)^2=9[/tex].
[tex]f(x)=-(x^2+6x+9-9+5)[/tex]
[tex]f(x)=-(x^2+2(x)(3)+3^2-4)[/tex]
[tex]f(x)=-(x^2+2(x)(3)+3^2)+4[/tex]
[tex]f(x)=-(x+3)^2+4[/tex]
On comparing this equation with [tex]f(x)=a(x-h)^2+k[/tex], we get
[tex]a=-1[/tex], it means the graph of the function is a downward parabola and the vertex is the point of maxima.
[tex]h=-3[/tex]
[tex]k=4[/tex]
The vertex of the function is (-3,4). So, the value of the function cannot be greater than 4.
Therefore, the range of the function is all real numbers less than or equal to 4.
Note: All options are incorrect.