The number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3
<h3>How to determine the number of real zeros?</h3>
The equation of the function is given as:

Expand the function

Reorder the terms

Factor the expression

Factor out x -1

Expand

Factorize
](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Bx%28x%20%2B%203%29%20%2B%202%28x%20%2B%203%29%5D%28x%20-%201%29)
Factor out x + 2

The function has been completely factored and it has 3 linear factors
Hence, the number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3
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Answer:
y= -1/4x + (-1)
Step-by-step explanation:
(4,-2) (0,-1)
m= y2-y1/x2-x1
m=-2-(-1)/4-0
m= -1/4
y= mx + b
-2= -1/4(4)+ b
-2= -1 +b
+1 +1
-1= b
3/8 times 2/5 is 6/40 or 0.15. Just multiply straight across
The answer is 130.8 hope it helps