Part A:
The y-intercept is $1,500
The x-intercept is 20 days
Part B:
Equation of the line in point-slope form is y - 1350 = -75×(x - 2)
Equation of the line in slope-intercept form is y = -75·x + 1500
Equation of the line in standard form is y + 75·x = 1500
Part C:
f(x) = -75·x + 1500
Part D:
$1,125
Explanation:
The information given are;
x, g(x)
0, $1,500
2, $1,350
4, $1,200
The equation of a straight line relation between the x, and y, coordinates in slope and intercept form is y = m·x + c
Where;
m = The slope
c = The y-intercept
From the data, we have two points, (x₁, y₁), (x₂, y₂) given as (0, 1,500), (2, 1350);
The y-intercept = y at x = 0 = 1,500
y = -75·x + 1500
x = 1500/75 = 20
Part B:
The equation in point-slope form is y - y₁ = m×(x - x₁)
Which gives;
Equation of the line in point-slope form is y - 1350 = -75×(x - 2)
Equation of the line in slope-intercept form is y = -75·x + 1500
Equation of the line in standard form is y + 75·x = 1500
Part C:
The equation of the line using function notation is f(x) = -75·x + 1500
Part D:
The balance in the bank after 5 days is f(5) = -75 × 5 + 1500 = $1,125