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write a piece-wise function for the following graph. Hint: find linear equations for each "piece" of the function with its associated domain restriction.

write a piece-wise function for the following graph. Hint: find linear equations for-example-1
User Cherona
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Answer

At values of x less than -2, (x < -2)

y = 1

At velues of x between -2 and 0, (-2 ≤ x < 0)

y = -1.5x - 2

At values of x between 0 and 3, (0 ≤ x < 3)

y = -2

At values of x between 3 and 5, (3 ≤ x < 5)

y = x - 5

At values of x greater than 5, (x > 5)

y = 3x - 15

Step-by-step explanation

To answer this, we will just break the domains into the x-value regions as obtainable from the graph. Starting from the left hand side

At values of x less than -2, (x < -2)

y = 1

At velues of x between -2 and 0, (-2 ≤ x < 0)

To write the equation of the line at this point, we need to note that

The slope and y-intercept form of the equation of a straight line is given as

y = mx + b

where

y = y-coordinate of a point on the line.

m = slope of the line.

x = x-coordinate of the point on the line whose y-coordinate is y.

b = y-intercept of the line.

For this question,

b = y-intercept = -2

We need to calculate the slope.

For a straight line, the slope of the line can be obtained when the coordinates of two points on the line are known. If the coordinates are (x₁, y₁) and (x₂, y₂), the slope is given as


Slope=m=\frac{Change\text{ in y}}{Change\text{ in x}}=(y_2-y_1)/(x_2-x_1)

For this question,

(x₁, y₁) and (x₂, y₂) are (-2, 1) and (0, -2)


\text{Slope = }(-2-1)/(0-(-2))=(-3)/(0+2)=(-3)/(2)=-1.5

So, for this region,

y = mx + b

y = -1.5x - 2

At values of x between 0 and 3, (0 ≤ x < 3)

y = -2

At values of x between 3 and 5, (3 ≤ x < 5)

We need to find the equation of the line at this point.

The general form of the equation in point-slope form is

y - y₁ = m (x - x₁)

where

y = y-coordinate of a point on the line.

y₁ = This refers to the y-coordinate of a given point on the line

m = slope of the line.

x = x-coordinate of the point on the line whose y-coordinate is y.

x₁ = x-coordinate of the given point on the line

Point = (x₁, y₁) = (3, -2)

x₁ = 3

y₁ = -2

So, we need to solve for the slope

(x₁, y₁) and (x₂, y₂) are (3, -2) and (5, 0)


\text{Slope = }(0-(-2))/(5-3)=(0+2)/(2)=(2)/(2)=1

y - y₁ = m (x - x₁)

m = 1

x₁ = 3, y₁ = -2

y - y₁ = m (x - x₁)

y - (-2) = 1 (x - 3)

y + 2 = x - 3

y = x - 3 - 2

y = x - 5

At values of x greater than 5, (x > 5)

For these parts, we just do it similarly to the region before this

Point = (x₁, y₁) = (5, 0)

x₁ = 5

y₁ = 0

(x₁, y₁) and (x₂, y₂) are (5, 0) and (6, 3)


\text{Slope = }(3-0)/(6-5)=(3)/(1)=3

y - y₁ = m (x - x₁)

m = 3

x₁ = 5, y₁ = 0

y - y₁ = m (x - x₁)

y - 0 = 3 (x - 5)

y = 3x - 15

Hope this Helps!!!

User Ajinkya Kulkarni
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