Given:
The equation is:

The graph of the
and
are given on a coordinate plane.
To find:
The solution of the given equation from the given graph.
Solution:
From the given graph it is clear that the graphs of
and
intersect each other at points (1.24,5.24) and (16,20).
It means the values of both functions
and
are equal at
and
.
So, the solutions of given equation are
and
.
Therefore, the correct option is only F.
The points where the 2 graphs intersect is where x = 0 and x = 2.
- 2 2
x = INT x dA / INT dA
0 0
INT dA = INT -x^2 + 4x + 3 - (x^2 + 3 ) dx = INT -2x^2 + 4x
= -2 x^3/3 + 2x^2
= 2.667 between 0 and 2
xdA = -2x^3 + 4x^2 INT xdA = -x^4/2 + 4x^3/3 = 2.667
centroid = 2.667 / 2.667 = 1 (x = 1)
Answer:
There are two x-intercepts.
A quadratic function whose maximum degree two. Therefore, Total two x-intercept possible.
We are given the range of quadratic function y less than equal to 2. It means parabola is downward whose maximum value 2.
Here y is greater than 2 and parabola form downward. Here must be two x-intercept.
If a>0 then form open up parabola.
If a<0 then open down parabola.
Here open down parabola. So, we get total two x-intercept.
Please see the attachment for diagram.
I would but I’m stuck on this problem too