Answer:
f(x) > 0 over the interval 
Step-by-step explanation:
If f(x) is a continuous function, and that all the critical points of behavior change are described by the given information, then we can say that the function crossed the x axis to reach a minimum value of -12 at the point x=-2.5, then as x increases it ascends to a maximum value of -3 for x = 0 (which is also its y-axis crossing) and therefore probably a local maximum.
Then the function was above the x axis (larger than zero) from 
, until it crossed the x axis (becoming then negative) at the point x = -4. So the function was positive (larger than zero) in such interval.
There is no such type of unique assertion regarding the positive or negative value of the function when one extends the interval from to -3, since between the values -4 and -3 the function adopts negative values.
The first case occurs in
for
and
. Extending the domain to account for all real
, we have this happening for
and
, where
.
The second case occurs in
when
, and extending to all reals we have
for
, i.e. any even multiple of
.
Answer:
this is the answer with steps
hope it helps!
I will assume that the two equations are
1) y = -x^2 +2x + 4 and 2) x +y = 4
subtract 2 from 1 and you get
y -x - y = -x^2 + 2x + 4 - 4
-x = -x^2 +2x
x^2 -x -2x = 0
x^2 - 3x = 0
x(x-3)=0
x=0 and x - 3 =0
x=0 and x = 3
Then y = 4 - x
y = 4 - 0 and y = 4 -3
y = 4 and y =1
Solutions:
Points (0,4) and (3,1)
