Problem solving with intergers
2 answers:
You might be interested in
Changing the equation into slope form: 
, where 
is the slope [gradient] and 
is the y-intercept.
The gradient is
and y-intercept is at 
Graphing using slope-intercept method:
a) The slope is a negative slope. The line will go 'down hill'
b) The line must pass the point (0, 490)
c) The line will intercept the x-axis at y = 0
So, x-intercept is at (735, 0)
The graph of this function is shown below. The intercepts are labelled at:
y = 490
x = 735
-----------------------------------------------------------------------------------------------------------
Next month's profit equation
Rewriting this into slope-equation form
The gradient,, equals to
The y-intercept,, equals to 531
The equation still has the same gradient with last month's profit equation but different y-intercept.
-------------------------------------------------------------------------------------------------------------
A linear graph show points of (0, 300) and (450, 0)
We work out the slope:
Y-intercept at x = 0, so it's at y = 300
Equation
The gradient is
Graphing
a) The slope is a negative slope. The line will go 'down hill'
b) The line must pass the point (0, 490)
c) The line will intercept the x-axis at y = 0
So, x-intercept is at (735, 0)
The graph of this function is shown below. The intercepts are labelled at:
y = 490
x = 735
-----------------------------------------------------------------------------------------------------------
Next month's profit equation
Rewriting this into slope-equation form
The gradient,
The y-intercept,
The equation still has the same gradient with last month's profit equation but different y-intercept.
-------------------------------------------------------------------------------------------------------------
A linear graph show points of (0, 300) and (450, 0)
We work out the slope:
Y-intercept at x = 0, so it's at y = 300
Equation