To solve for the [tex]\( x \)[/tex]-intercept of the continuous linear function [tex]\( k \)[/tex], we start with the provided information:
- The average rate of change (or the slope [tex]\( m \)[/tex]) is [tex]\(-3\)[/tex].
- The [tex]\( y \)[/tex]-intercept ([tex]\( b \)[/tex]) is [tex]\( 5 \)[/tex].
Linear functions can be expressed in the slope-intercept form [tex]\( y = mx + b \)[/tex]. Therefore, for our function [tex]\( k \)[/tex], the equation is:
[tex]\[ y = -3x + 5 \][/tex]
The [tex]\( x \)[/tex]-intercept is the point where the graph of the function crosses the [tex]\( x \)[/tex]-axis. At the [tex]\( x \)[/tex]-intercept, the value of [tex]\( y \)[/tex] is [tex]\( 0 \)[/tex]. This gives us the equation:
[tex]\[ 0 = -3x + 5 \][/tex]
To find the [tex]\( x \)[/tex]-intercept, we solve this equation for [tex]\( x \)[/tex]:
1. Set up the equation:
[tex]\[ 0 = -3x + 5 \][/tex]
2. Add [tex]\( 3x \)[/tex] to both sides to isolate the variable term:
[tex]\[ 3x = 5 \][/tex]
3. Divide both sides by [tex]\( 3 \)[/tex]:
[tex]\[ x = \frac{5}{3} \][/tex]
Hence, the [tex]\( x \)[/tex]-intercept of the function [tex]\( k \)[/tex] is at:
[tex]\[ x = \frac{5}{3} \][/tex]
As a decimal, this is approximately:
[tex]\[ x \approx 1.6666666666666667 \][/tex]
So, the [tex]\( x \)[/tex]-intercept of the function [tex]\( k \)[/tex] is at [tex]\( \frac{5}{3} \)[/tex].
