A. ) f(x) has one real zero at –2 because the graph of the function has an intercept at (0, –2).
B. ) f(x) has two real zeros at –4 and –2 because the graph of the function has intercepts at (–4, 0) and (0, –2).
C. ) f(x) has no real zeros because the graph of the function does not pass through (0, 0).
D. ) f(x) has one real zero at –4 because the graph of the function has an intercept at (–4, 0).
Answer:
C. ) f(x) has no real zeros because the graph of the function does not pass through (0, 0).
Step-by-step explanation:
According to L'Hospital's Rule, this is true.
Step-by-step explanation:
HOPE IT HELPS Y
y = –3x + 11
A.
(6, -3)
B.
(6, -7)
D.
(5, -4)
C:
Hi Brainiac
3x+2y=7
y=-3x+11
We need to solve y=-3x+11 for y
Now let's start by substitute -3x+11 for y in 3x+2y=7
3x+2y=7
3x+2(-3x+11)=7
-3x+22=7
Now add -22 to both sides
-3x+22-22=7-22
-3x=-15
To find the value for x we need to divide both sides by -3
-3x/-3= -15/-3
x=5
Now we have the value for x :)
Let's find the value for y by substitute 5 for x in y=-3x+11
y=-3x+11
y=-3(5)+11
y=-4
Here you go :)
The answer is D
Good luck :0