Find the indicated limit, if it exists. limit of f of x as x approaches negative 1 where f of x equals x plus 1 when x is less t
han negative 1 and f of x equals 1 minus x when x is greater than or equal to negative 1
1 answer:
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Answer:
see explanation
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Given
- 4x + 3y = 12 ( add 4x to both sides )
3y = 4x + 12 ( divide the terms by 3 )
y = x + 4 ← in slope- intercept form
with slope m = and y- intercept c = 4
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Given
- 5x + 3y = - 9 ( add 5x to both sides )
3y = 5x - 9 ( divide the terms by 3 )
y = x - 3 ← in slope- intercept form
with slope m = and y- intercept c = - 3
Answer:
Domain = ( -∞,∞), {x|x ∈ R}
Range (-∞,2], {y|y ≤ 2}
Vertex (h,k) = (6,2)
Step-by-step explanation:
(Domain / Range) The absolute value expression has a V shape. The range of a negative absolute value expression starts at its vertex and extends to negative infinity.
(Vertex) To find the x coordinate of the vertex, set the inside of the absolute value
x − 6 equal to 0 . In this case, x − 6 = 0 .
x−6=0
Add 6 to both sides of the equation.
x=6
Replace the variable x with 6 in the expression.
y=−1/3⋅|(6)−6|+2
Simplify−1/3⋅|(6)−6|+2.
y=2
The absolute value vertex is ( 6 , 2 ) .
(6,2)
Hope this helps
