a + 5 = 2
a + 5 - 5 = 2 - 5
a = -3
Hope this helps! :)
Find all relative extrema of the function. use the second derivative test where applicable. (if an answer does not exist, enter
dne.) f(x) = x2 + 3x − 1
1 answer:
The given quadratic describes a parabola that opens upward. Its one absolute extreme is a minimum that is found at x = -3/2. The value of the function there is
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at (-1.5, -3.25).
_____
For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at (-1.5, -3.25).
_____
For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.
You might be interested in
We have
f(x)=<span>−x</span>²<span>+ 5x + 6
g(x)=</span><span>3x − 2
we know that
f(x)=g(x)
the solution is the intersection both graphs
using a graph tool
see the attached figure
the solution are the points
(4,10)
and
(-2,-8)
</span>
f(x)=<span>−x</span>²<span>+ 5x + 6
g(x)=</span><span>3x − 2
we know that
f(x)=g(x)
the solution is the intersection both graphs
using a graph tool
see the attached figure
the solution are the points
(4,10)
and
(-2,-8)
</span>