Peer Response 2

1. Respond to your classmates’ posts in at least 25 words.

PEERS RESPONSE:

Hello class! Hope everyone is enjoying their week.

 

Problem 1

For my first problem (#20 on page 708), I will graph each absolute value function and state the domain y = |x – 1| + 2.

Step 1: We will begin to solve for various x values needed to graph the absolute value function, subbing f(x) for y.

f(x) = |x – 1| + 2

f(-2) = |-2 – 1| + 2 = 3 + 2 = 5, which translates to the coordinate (-2, 5)

f(-1) = |-1 – 1| + 2 = 2 + 2 = 4, which translates to the coordinate (-1, 4)

f(0) = |x – 1| + 2 = 1 +2 = 3, which translates to the coordinate (0, 3)

f(1) = |1 – 1| + 2 = 0 + 2 = 2, which translates to the coordinate (1, 2)

f(2) = |2 – 1| + 2 = 1 + 2 = 3, which translates to the coordinate (2, 3)

Step 2:  We plotted the 5 points above on a graph, providing us with the following results:

· This absolute value function is a V-shaped graph that opens upward, extending to the left and right without restriction.

· The equation is a function because it passes the vertical line test, meaning there is only one y value for every x value.

· Since this is an absolute value function, the value of x cannot be negative; therefore, the graph remains above the x-axis and does not have an x-intercept. The y-intercept is (0, 3).

· The domain of the equation in interval notation is (-∞, ∞) and the range is [2, ∞).

Step 3: If we were to take this same function and shift it up 3 units and left 4 units, this equation will be affected by:

· Adding 3 outside of the absolute value bars to shift the function upward 3 units.

· Adding 4 inside the absolute value bard to shift the function left 4 units.

Step 4: This means the equation now looks like y = |x + 3| + 5, reflecting the transformation of the absolute value function up 3 units and left 4 units.

Problem 2

For my next problem (#28 on page 721), I will graph the function f(x) = x2 + 2

Step 1: We will begin to solve for various x values needed to graph the function f(x) = x2 + 2

f(-2) = -22 + 2 = 4 + 2 = 6, which translates to the coordinate (-2, 6)

f(-1) = -12 + 2 = 1 + 2 = 3, which translates to the coordinate (-1, 3)

f(0) = 02 + 2 = 0 + 2 = 2, which translates to the coordinate (0, 2)

f(1) = 12 + 2 = 1 + 2 = 3, which translates to the coordinate (1, 3)

f(2) = 22 + 2 = 4 + 2 = 6, which translates to the coordinate (2, 6)

 

Step 2:  We plotted the 5 points above on a graph, providing us with the following results:

· This function is shaped like a parabola that opens upward, extending to the left and right without restriction.

· The equation is a function because it passes the vertical line test, meaning there is only one y value for every x value.

· Since the value of x is squared within the function, the value of x cannot be negative; therefore, the graph remains above the x-axis and does not have an x-intercept. The y-intercept is (0, 2).

· The domain of the equation in interval notation is (-∞, ∞) and the range is [2, ∞).

 

References:

Dugopolski, M. (2012).  Elementary and intermediate algebra (4th ed.). McGraw-Hill Education.

-Kyrstal