WRITE A PIECEWISE FUNCTION FOR THE GIVEN GRAPH

Write the function represented by the graphs given below.

Problem 1 :

Solution :

The graph, above the x-axis is a raising line, it must be a linear function, then it must be in the form of y = mx + b

Equation of the line above x-axis :

Slope of the line = 3/1 ==> 3

y = 3x + b

Applying one of the points on the line into this equation above, we find the value of b.

(3, 9) is one of the points on the line

9 = 3(3) + b

b = 9 - 9

b = 0

So, the required function for the curve above x-axis is y = 3x.

Equation of the line below x-axis :

Slope of the line = 2/2 ==> 1

y-intercept = -4

y = x - 4

Problem 2 :

Solution :

From left to right, the first line is falling line.

Slope = -2/1 ==> -2

y-intercept = -1

y = -2x - 1

Domain for this function is x ≤ 2

From the left to right the second line is also falling line.

Slope = -1/1 ==> -1

y = -1x + b ---(1)

Applying one of the points on the line.

(4, 0)

0 = -1(4) + b

0 = -4 + b

b = 4

y = -x + 4

Domain for this function is x > 2

Problem 3 :

Solution :

We have three parts in the piecewise function.

1) Horizontal line

2) Raising line

3) Falling line

Equation of horizontal line :

y = -4

Domain for this function is x ≤ -2

Equation of raising line :

Slope = 1, y-intercept = -2

y = x - 2

Domain for this function is -2 < x < 2

Equation of falling line :

Slope = -2/1 => -2

y = -2x + b

Applying one of the points on the line.

The line is passing through the point (4, -4).

-4 = -2(4) + b

-4 = -8 + b

b = -4 + 8

b = 4

y = -2x + 4

Domain for this function is x ≥ 2

Problem 4 :

Solution :

One piece can be considered as a absolute value function.

Vertex of the absolute value function is (0,0) and it opens down.

y = a|x - h| + k

y = a|x - 0| + 0

y = a |x|

Applying one of the points on the absolute value function to figure out the value of a, we get

By applying the point (-2, -2)

-2 = a|-2|

-2 = 2a

a = -1

So, the required absolute value function is y = -|x| and its domain is -4 ≤ x ≤ 0

The other part is parabola,

y = a(x - h)² + k

y = a(x - 0)² + 4

y = ax ²+ 4

The curve is passing through the point (2, 0).

0 = a(2)² + 4

4a = -4

a = -1

By applying the value of a, we get

y = -x²+ 4

It's domain is 0 < x < ∞

Problem 5 :

Solution :

One part is the graph of reciprocal function. It's vertical asymptote is at x = 4.

Since the horizontal asymptote is x-axis o y = 0. The highest exponent of denominator is greater than the numerator.

So, the required function is f(x) = 1/(x - 4) and its domain is x ≤ 4.

Another part is horizontal line. y = -4 and its domain is 4 < x < ∞

Problem 6 :

Solution :

The three parts are horizontal line, falling line and raising line.

Equation of horizontal line :

y = 2 and its domain is x < -4

Equation of falling line :

Slope = -1 and y-intercept = -2

y = -x - 2 and its domain is -4 < x < 2

Equation of falling line :

Slope = 1

y = x + b

To find the value of b, applying one of the point lines on the line into the function.

The line is passing through the point (4, -2).

-2 = 4 + b

b = -6

Applying the value of b into the function, we get y = x - 6. It's domain is x > 2