Question

write an equation for a function that has a rate of change equal to \\(\frac{13}{2}\\), and a starting point of 5.\
y = \\(\square\\) x + \\(\square\\)\
question 8\
8 pts\
write an equation for a function that has a rate of change equal to \\(-\frac{2}{7}\\), and a starting point of -1.\
y = \\(\square\\) x + \\(\square\\)

Explanation:

Step1: Recall slope-intercept form

The slope - intercept form of a linear function is \(y = mx + b\), where \(m\) is the rate of change (slope) and \(b\) is the starting point (y - intercept).

Step2: Solve the first equation

For the first function, the rate of change \(m=\frac{13}{2}\) and the starting point \(b = 5\). So substituting into \(y=mx + b\), we get \(y=\frac{13}{2}x+5\).

Step3: Solve the second equation

For the second function, the rate of change \(m =-\frac{2}{7}\) and the starting point \(b=- 1\). So substituting into \(y = mx + b\), we get \(y=-\frac{2}{7}x-1\).

Answer:

First equation: \(y=\boldsymbol{\frac{13}{2}}x+\boldsymbol{5}\)

Second equation: \(y=\boldsymbol{-\frac{2}{7}}x+\boldsymbol{(-1)}\) (or \(y =-\frac{2}{7}x - 1\))