The problem gives us a line [tex]\( f \)[/tex] with a slope of [tex]\(-\frac{9}{2}\)[/tex] and that passes through the point [tex]\((0, 15)\)[/tex]. We need to determine the [tex]\( x \)[/tex]-intercept of this line.
To start, we can use the slope-intercept form of a line, which is given by:
[tex]\[
y = mx + b
\][/tex]
Here, [tex]\( m \)[/tex] is the slope of the line, and [tex]\( b \)[/tex] is the y-intercept. According to the problem, the slope [tex]\( m \)[/tex] is [tex]\(-\frac{9}{2}\)[/tex], and the line passes through the point [tex]\((0, 15)\)[/tex].
Since [tex]\((0, 15)\)[/tex] is where the line intersects the y-axis, [tex]\( b \)[/tex] is 15. Inserting these values into the slope-intercept form, we get:
[tex]\[
y = -\frac{9}{2}x + 15
\][/tex]
Next, we need to find the [tex]\( x \)[/tex]-intercept, which is the point where the line crosses the [tex]\( x \)[/tex]-axis. The [tex]\( x \)[/tex]-intercept occurs where [tex]\( y = 0 \)[/tex]. Setting [tex]\( y \)[/tex] to 0 in the equation of the line, we have:
[tex]\[
0 = -\frac{9}{2}x + 15
\][/tex]
To solve for [tex]\( x \)[/tex], we first isolate the term involving [tex]\( x \)[/tex]. Adding [tex]\(\frac{9}{2}x\)[/tex] to both sides of the equation, we obtain:
[tex]\[
\frac{9}{2}x = 15
\][/tex]
Next, to solve for [tex]\( x \)[/tex], we multiply both sides by the reciprocal of [tex]\(\frac{9}{2}\)[/tex], which is [tex]\(\frac{2}{9}\)[/tex]:
[tex]\[
x = 15 \times \frac{2}{9}
\][/tex]
Carrying out the multiplication, we get:
[tex]\[
x = \frac{30}{9} = \frac{10}{3}
\][/tex]
Hence, the [tex]\( x \)[/tex]-intercept of the line [tex]\( f \)[/tex] is [tex]\(\frac{10}{3}\)[/tex], which is approximately:
[tex]\[
x \approx 3.3333333333333335
\][/tex]
