Answer:
Zeros of the given function are x=5 and x=-1.
Explanation:
f(x)=x^2-4x-5
f(x)=x^2+1x-5x-5
f(x)=x(x+1)-5(x+1)
f(x)=(x-5)(x+1)
To find zeros, we need to set f(x)=0
0=(x-5)(x+1)
0=(x-5) or 0=(x+1)
0=x-5 or 0=x+1
5=x or -1=x
Hence zeros of the given function are x=5 and x=-1.
We can plug some random numbers like x=0,1,2,... into given function to find few points then graph those points and join them by a curved line.
That will give the final graph as attached below:
for x=0,
f(x)=x^2-4x-5
f(0)=0^2-4(0)-5
f(0)=0-0-5
f(0)=-5
Hence first point is (0,-5)
Similarly we can find more points.