Given:
Consider the given function is [tex]f(x)=-(x+1)(x-3)(x+2)[/tex].
To find:
The remaining zero and y-coordinate of y-intercept.
Solution:
We have,
[tex]f(x)=-(x+1)(x-3)(x+2)[/tex]
For zeros, [tex]f(x)=0[/tex].
[tex]-(x+1)(x-3)(x+2)=0[/tex]
[tex](x+1)(x-3)(x+2)=0[/tex]
[tex](x+1)=0,(x-3)=0,(x+2)=0[/tex]
[tex]x=-1,x=3,x=-2[/tex]
So, three zeros of given function are -1, 3 and -2.
Putting x=0 in the given function, we get
[tex]f(0)=-(0+1)(0-3)(0+2)[/tex]
[tex]f(0)=-(1)(-3)(2)[/tex]
[tex]f(0)=-(-6)[/tex]
[tex]f(0)=6[/tex]
So, the y-coordinate of y-intercept of the given function is 6. It means the y-intercept is at point (0,6).
Therefore, the zeros of the function [tex]f(x)=-(x+1)(x-3)(x+2)[/tex] are –1, 3, and -2 and the y-intercept of the function is located at (0,6).
