Transcript 2.3 - Mathematics page

Rate of change and slope
r= d/t
r
fx = x
d
r=
rise/run
4
2
5
10
t
Rate of change =
Change in y /change in x
Or
Rate of change =Δy/Δx
Rate of change = rise / run
15
The slope of the line is the ratio of the change in
y-coordinates to the change in the xcoordinates
OR
m =( y2 – y1 ) / ( x2 – x1 )
Δy
where m is the slope
Δx
Find the slope of the line passing through the
following points : ( -1.5 , 3.5 ) and ( 4.5 , 6)
m=( y2 – y1 ) / ( x2 - x 1 )
m=
= 5/ 12
6
gx = 2x+3
( 1,5), (-1 , 1)
4
2
5
10
m= (1-5) /( -1-1 ) = -4 / -2
Δy
Δx
m=2
15
hx = -2x+6
8
6
( -1 , 8 ) and ( 2 , 2 )
m= ( 2 – 8 )/(2 – (-1 ))
m = -6 / 3
m= -2
4
2
5
-2
10
15
fx = 5
10
8
6
4
2
5
There is run
but No rise
10
15
m = o/ 2 or
0/5
m=0
20
fy = 3
m = 5/ 0 or
7/0
m is undefined
10
8
6
4
2
5
10
15
20
Slope is
undefined
Slope is zero
Example :
Find the rate of change for each equation
1) 6y = 8x – 40 ( ÷ 2 )
3y = 4x -20
(÷3)
y = 4/3 x – 20/3
m = 4/3
2 ) -2y -16 x = 41
-2 y = 16 x + 41
y = -8x – 41/ 2
m=-8
(÷- 2)
3) 12 x – 4y + 5 =18
12 x – 4y = 13
12 x – 13 = 4 y
(÷ 4 )
3 x – 13 / 4 = y
m=3
4)
Multiply each term by 4
6x – 5 y = 60
6 x – 60 = 5 y
6/5 x - 12 = y
m = 6/5
Unknown value
Find the value of r that the line passes through each pair
of points has the given slope.
(10,r) ,(4,3) , m = 4/3
m =( y2 – y1 ) / ( x2 – x1 )
m= (3-r)/(4-10) = ( 3 – r ) / - 6
r= 11
Required exercises are :
3, 9, 10, 11(a, b, c), 12 to 21, 23 to
28, 31 to 34, 36, 37, 40(a, b, c),
41, 42, 43, 44.
Pages : 79-82
Thank you