Question

Type the correct answer in each box. Use numerals instead of words.

Answer 1

Given:

Consider the given function is $f(x)=-(x+1)(x-3)(x+2)$.

To find:

The remaining zero and y-coordinate of y-intercept.

Solution:

We have,

$f(x)=-(x+1)(x-3)(x+2)$

For zeros, $f(x)=0$.

$-(x+1)(x-3)(x+2)=0$

$(x+1)(x-3)(x+2)=0$

$(x+1)=0,(x-3)=0,(x+2)=0$

$x=-1,x=3,x=-2$

So, three zeros of given function are -1, 3 and -2.

Putting x=0 in the given function, we get

$f(0)=-(0+1)(0-3)(0+2)$

$f(0)=-(1)(-3)(2)$

$f(0)=-(-6)$

$f(0)=6$

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 $f(x)=-(x+1)(x-3)(x+2)$ are –1, 3, and -2 and the y-intercept of the function is located at (0,6).