How to algebraically find x and y intercepts

Solution

We want the distance between xxand 5 to be less than or equal to 4. We can draw a number line, such as the one in Figure 2, to represent the condition to be satisfied.

Figure 2

The distance from xxto 5 can be represented using the absolute value as |x−5|.|x−5|.We want the values of xxthat satisfy the condition |x−5|≤4.|x−5|≤4.

Solution

5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance should not exceed the stated variability, so, with the resistance RRin ohms,

|R−680|≤34|R−680|≤34

Solution

The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function. See Figure 6.

Figure 6

We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. Instead, the width is equal to 1 times the vertical distance as shown in Figure 7.

Figure 7

From this information we can write the equation

f(x)=2|x−3|−2,treating the stretch as a vertical stretch, orf(x)=|2(x−3)|−2,treating the stretch as a horizontal compression.f(x)=2|x−3|−2,treating the stretch as a vertical stretch, orf(x)=|2(x−3)|−2,treating the stretch as a horizontal compression.

Solution

0=|4x+1|−70=|4x+1|−7Substitute 0 for f(x).7=|4x+1|7=|4x+1|Isolate the absolute value on one side of the equation.7=4x+1or−7=4x+16=4x−8=4xx=64=1.5   x=−84=−27=4x+1or−7=4x+16=4x−8=4xx=64=1.5   x=−84=−2Break into two separate equations and solve.

The function outputs 0 when x=1.5x=1.5or x=−2.x=−2.See Figure 9.

Figure 9

Solution

Isolating the absolute value on one side of the equation gives the following.

1=4|x−2|+2−1=4|x−2|−14=|x−2|1=4|x−2|+2−1=4|x−2|−14=|x−2|

The absolute value always returns a positive value, so it is impossible for the absolute value to equal a negative value. At this point, we notice that this equation has no solutions.

Solution

With both approaches, we will need to know first where the corresponding equality is true. In this case we first will find where |x−5|=4.|x−5|=4.We do this because the absolute value is a function with no breaks, so the only way the function values can switch from being less than 4 to being greater than 4 is by passing through where the values equal 4. Solve |x−5|=4.|x−5|=4.

x−5=4x=9orx−5=−4x=1x−5=4x=9orx−5=−4x=1

After determining that the absolute value is equal to 4 at x=1x=1and x=9,x=9,we know the graph can change only from being less than 4 to greater than 4 at these values. This divides the number line up into three intervals:

x<1,1<x<9, and  x>9.x<1,1<x<9, and  x>9.

To determine when the function is less than 4, we could choose a value in each interval and see if the output is less than or greater than 4, as shown in Table 1.

Interval test xxf(x)f(x)<4<4 or >4?>4?x<1x<10|0−5|=5|0−5|=5Greater than1<x<91<x<96|6−5|=1|6−5|=1Less thanx>9x>911|11−5|=6|11−5|=6Greater than

Table 1

Because 1<x<91<x<9is the only interval in which the output at the test value is less than 4, we can conclude that the solution to |x−5|<4|x−5|<4is 1<x<9,1<x<9,or (1,9).(1,9).

To use a graph, we can sketch the function f(x)=|x−5|.f(x)=|x−5|.To help us see where the outputs are 4, the line g(x)=4g(x)=4could also be sketched as in Figure 11.

Figure 11 Graph to find the points satisfying an absolute value inequality.

We can see the following:

• The output values of the absolute value are equal to 4 at x=1x=1 and x=9.x=9.
• The graph of ff is below the graph of gg on 1<x<9.1<x<9. This means the output values of f(x)f(x) are less than the output values of g(x).g(x).
• The absolute value is less than or equal to 4 between these two points, when 1<x<9.1<x<9. In interval notation, this would be the interval (1,9).(1,9).

Solution

We are trying to determine where f(x)<0,f(x)<0,which is when −12|4x−5|+3<0.−12|4x−5|+3<0.We begin by isolating the absolute value.

−12|4x−5|<−3Multiply both sides by –2, and reverse the inequality.|4x−5|>6−12|4x−5|<−3Multiply both sides by –2, and reverse the inequality.|4x−5|>6

Next we solve for the equality |4x−5|=6.|4x−5|=6.

4x−5=6  or4x−5=−64x−5=64x=−1x=114x=−144x−5=6  or4x−5=−64x−5=64x=−1x=114x=−14

Now, we can examine the graph of ffto observe where the output is negative. We will observe where the branches are below the x-axis. Notice that it is not even important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at x=−14x=−14and x=114x=114and that the graph has been reflected vertically. See Figure 12.

Figure 12

We observe that the graph of the function is below the x-axis left of x=−14x=−14and right of x=114.x=114.This means the function values are negative to the left of the first horizontal intercept at x=−14,x=−14,and negative to the right of the second intercept at x=114.x=114.This gives us the solution to the inequality.

x<−14 or x>114x<−14 or x>114

In interval notation, this would be (−∞,−0.25)∪(2.75,∞).(−∞,−0.25)∪(2.75,∞).