Answer :
To find the [tex]$x$[/tex]-intercepts of the function [tex]\( f(x) = 3 |x + 3| - 7 \)[/tex], we need to determine where the graph of the function crosses the x-axis. This occurs when [tex]\( f(x) = 0 \)[/tex]. So, we need to solve the equation:
[tex]\[ 3|x + 3| - 7 = 0 \][/tex]
First, we isolate the absolute value term:
[tex]\[ 3|x + 3| = 7 \][/tex]
Next, we divide both sides of the equation by 3:
[tex]\[ |x + 3| = \frac{7}{3} \][/tex]
The absolute value equation [tex]\( |x + 3| = \frac{7}{3} \)[/tex] means that [tex]\( x + 3 \)[/tex] can be either [tex]\( \frac{7}{3} \)[/tex] or [tex]\( -\frac{7}{3} \)[/tex]. Therefore, we need to consider two cases:
1. [tex]\( x + 3 = \frac{7}{3} \)[/tex]
2. [tex]\( x + 3 = -\frac{7}{3} \)[/tex]
Case 1:
[tex]\[ x + 3 = \frac{7}{3} \][/tex]
Subtract 3 from both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{7}{3} - 3 \][/tex]
Convert the 3 to a fraction with a denominator of 3:
[tex]\[ x = \frac{7}{3} - \frac{9}{3} \][/tex]
Subtract the fractions:
[tex]\[ x = \frac{7 - 9}{3} \][/tex]
[tex]\[ x = \frac{-2}{3} \][/tex]
[tex]\[ x = -\frac{2}{3} \][/tex]
Case 2:
[tex]\[ x + 3 = -\frac{7}{3} \][/tex]
Subtract 3 from both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = -\frac{7}{3} - 3 \][/tex]
Convert 3 to a fraction with a denominator of 3:
[tex]\[ x = -\frac{7}{3} - \frac{9}{3} \][/tex]
Subtract the fractions:
[tex]\[ x = \frac{-7 - 9}{3} \][/tex]
[tex]\[ x = \frac{-16}{3} \][/tex]
[tex]\[ x = -\frac{16}{3} \][/tex]
Therefore, the [tex]\( x \)[/tex]-intercepts of the function [tex]\( f(x) = 3 |x + 3| - 7 \)[/tex] are at:
[tex]\[ x = -\frac{2}{3}, -\frac{16}{3} \][/tex]
So, the intercepts are at [tex]\( x = -\frac{2}{3}, -\frac{16}{3} \)[/tex].
[tex]\[ 3|x + 3| - 7 = 0 \][/tex]
First, we isolate the absolute value term:
[tex]\[ 3|x + 3| = 7 \][/tex]
Next, we divide both sides of the equation by 3:
[tex]\[ |x + 3| = \frac{7}{3} \][/tex]
The absolute value equation [tex]\( |x + 3| = \frac{7}{3} \)[/tex] means that [tex]\( x + 3 \)[/tex] can be either [tex]\( \frac{7}{3} \)[/tex] or [tex]\( -\frac{7}{3} \)[/tex]. Therefore, we need to consider two cases:
1. [tex]\( x + 3 = \frac{7}{3} \)[/tex]
2. [tex]\( x + 3 = -\frac{7}{3} \)[/tex]
Case 1:
[tex]\[ x + 3 = \frac{7}{3} \][/tex]
Subtract 3 from both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{7}{3} - 3 \][/tex]
Convert the 3 to a fraction with a denominator of 3:
[tex]\[ x = \frac{7}{3} - \frac{9}{3} \][/tex]
Subtract the fractions:
[tex]\[ x = \frac{7 - 9}{3} \][/tex]
[tex]\[ x = \frac{-2}{3} \][/tex]
[tex]\[ x = -\frac{2}{3} \][/tex]
Case 2:
[tex]\[ x + 3 = -\frac{7}{3} \][/tex]
Subtract 3 from both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = -\frac{7}{3} - 3 \][/tex]
Convert 3 to a fraction with a denominator of 3:
[tex]\[ x = -\frac{7}{3} - \frac{9}{3} \][/tex]
Subtract the fractions:
[tex]\[ x = \frac{-7 - 9}{3} \][/tex]
[tex]\[ x = \frac{-16}{3} \][/tex]
[tex]\[ x = -\frac{16}{3} \][/tex]
Therefore, the [tex]\( x \)[/tex]-intercepts of the function [tex]\( f(x) = 3 |x + 3| - 7 \)[/tex] are at:
[tex]\[ x = -\frac{2}{3}, -\frac{16}{3} \][/tex]
So, the intercepts are at [tex]\( x = -\frac{2}{3}, -\frac{16}{3} \)[/tex].
