FIND X AND Y INTERCEPTS OF ABSOLUTE VALUE FUNCTION

x-intercept :

The curve where it intersects the x-axis is known as x-intercept. To find x-intercept, we put y = 0.

y-intercept :

The curve where it intersects the y-axis is known as y-intercept. To find y-intercept, we put x = 0.

Find x and y intercepts for the absolute value function given below.

Problem 1 :

y = 3|x - 1| + 2

Solution :

x -Intercept:

put y = 0

3|x - 1| + 2 = 0

3|x - 1| = -2

|x - 1| = -2/3

This is not admissible. There is no real solution. So, there is no x-intercept. 

y -Intercept:

Put x = 0

y = 3|0 - 1| + 2

y = 3(1) + 2

y = 3 + 2

y = 5

y -Intercept is (0, 5).

Problem 2 :

y = 2|x|

Solution :

x -Intercept:

Put y = 0

2|x| = 0

x = 0

x -Intercept is (0, 0).  

y -Intercept:

Put x = 0

y = 2|0|

y = 0

y -Intercept is (0, 0).

Problem 3 :

y = |x| + 5

Solution :

x -Intercept :

Put y = 0

|x| + 5 = 0

|x| = - 5

The given function will not intersect the x-axis.

y -Intercept :

put x = 0

y = 0 + 5

y = 5

y -Intercept is (0, 5).

Problem 4 :

y = -2|x + 1| - 3

Solution :

x -Intercept:

Put y = 0

-2|x + 1| - 3 = 0

-2|x + 1| = 3

|x + 1| = -3/2

There is no x-intercept.

y -Intercept :

put x = 0

y = -2|0 + 1| - 3

= -2(1) - 3

= -2 - 3

y = -5

y -Intercept is (0, -5)   

Problem 5 :

y = -3/5|x + 3| + 10

Solution :

x -Intercept:

 put y = 0

-3/5|x + 3| + 10 = 0

-3/5|x + 3| = -10

3/5|x + 3| = 10

|x + 3| = 50/3

So, x-intercepts are (41/3, 0) and (-59/3, 0).

y -Intercept :

put x = 0

y = -3/5|0 + 3| + 10

y = -3/5(3) + 10

y = -9/5 + 10

y = (-9 + 50)/5

y = 41/5

y -Intercept is (0, 41/5)   

Problem 6 :

y = 15|x|

Solution :

x -Intercept :

put y = 0

15|x| = 0

x = 0

x -Intercept is (0, 0).

y -Intercept :

put x = 0

y = 15(0)

y = 0

y -Intercept is (0, 0).

Problem 7 :

y = 5/3|x + 2| - 1

Solution :

x -Intercept:

put y = 0

5/3|x + 2| - 1 = 0

5/3|x + 2| = 1

|x + 2| = 3/5

x -Intercept are (-7/5, 0) and (-13/5, 0).

y -Intercept :

put x = 0

y = 5/3|0 + 2| - 1

y = 5/3(2) - 1

y = 10/3 - 1

y = 7/3

y -Intercept (0, 7/3).

Problem 8 :

y = -2|x + 1| - 3

Solution :

x -Intercept :

put y = 0

-2|x + 1| - 3 = 0

-2|x + 1| = 3

|x + 1| = -3/2

There is no x-intercept.

y -Intercept :

put x = 0

y = -2|0 + 1| - 3

y = -2(1) - 3

y = -2 - 3

y = -5

y -Intercept is (0, -5).

Problem 9 :

You ride your bicycle around a circular trail one time. The function

f(x) = (−1/3) ∣x − 4.5∣ + 1.5

represents the shortest distance (in miles) along the trail between you and your starting point after x minutes.

(a) Graph the function. Find the domain and range in this context.

(b) Interpret the intercepts and the vertex.

Solution :

f(x) = (−1/3) ∣x − 4.5∣ + 1.5

a. The function is in vertex form, g(x) = a ∣ x − h ∣ + k, where h = 4.5 and k = 1.5. So, the vertex is (h, k) = (4.5, 1.5). Substitute 0 for f (x) to find any x-intercepts.

0 = (−1/3) ∣x − 4.5∣ + 1.5

-1.5 = (−1/3) ∣x − 4.5∣

1.5(3) = ∣x − 4.5∣

∣x − 4.5∣ = 4.5

Graph the function using the points (0, 0), (4.5, 1.5), and (9, 0). The time x and distance f ( x) must be greater than or equal to 0 in this context. So, the domain is {x | 0 ≤ x ≤ 9} and the range is { f (x) | 0 ≤ f (x) ≤ 1.5}.

b. The intercepts 0 and 9 indicate that it takes you 9 − 0 = 9 minutes to ride around the trail. The vertex (4.5, 1.5) indicates that you are 1.5 miles from your starting point after 4.5 minutes. You can also determine that you are halfway around the trail at that time.