Given :
On a coordinate plane,a curved line with 3 arcs, lab led f of x, crosses the x-axis at (negative 2,0), (negative 1,0), (1,0), and (3,0) and the y axis at (0, negative 6).
To find:
f when x = 0. i.e. f (0).
Solution:
since the graph has 3 arcs and 4 solutions, it can be visualized as the follows:
Between each solution, the function has to increase and decrease giving arcs in between.
1. One of the arcs is between (negative 2,0) and negative (1,0)
2. Second arc is between (negative 1,0) and (1,0)
-this arc cuts the y axis, since x= 0 lies between x= -1 & x=1-
3. Third arc is between (1,0) and (3,0)
Therefore only the 2nd arc cuts the y axis
It’s given that the curve cuts the y axis at (0, -6)
That is when x= 0, f(0) =-6
Therefore the value of f (0) is -6 only.
HOPED THIS HELPED LUV!!
Answer:
Formular for coordinate length:
In considerance of coordinates given of A and B
therefore substitute:
The number of whole number is 23 through subtracting the numbers. hope it helps ☺️
X + 2y - 6 = z..........x + 2y - z = 6
3y - 2z = 7
4 + 3x = 2y - 5z.....3x - 2y + 5z = -4
x + 2y - z = 6.....multiply by -3
3x - 2y + 5z = -4
---------------------
-3x - 6y + 3z = -18 (result of multiplying by -3)
3x - 2y + 5z = -4
--------------------add
-8y + 8z = - 22
-8y + 8z = -22
3y - 2z = 7 .....multiply by 4
----------------
-8y + 8z = -22
12y - 8z = 28 (result of multiplying by 4)
----------------add
4y = 6
y = 6/4 reduces to 3/2
-8y + 8z = -22
-8(3/2) + 8z = -22
-12 + 8z = -22
8z = -22 + 12
8z = -10
z = -10/8 reduces to -5/4
x + 2y - 6 = z
x + 2(3/2) - 6 = -5/4
x + 3 - 6 = -5/4
x - 3 = -5/4
x = -5/4 + 3
x = -5/4 + 12/4
x = 7/4
solution is (7/4, 3/2, -5/4).....x = 7/4, y = 3/2, and z = -5/4
