Function G has slope 1/3 and passes through (0, -1). Function H passes through (8, 3) and (10, 4). Kelly says the initial value of G is greater than the initial value of H. Is Kelly correct?

Step-by-step solution

1. 1

   Define 'initial value'. For a linear function, the initial value is the output when the input is $0$ — that is, the $y$-intercept. So the whole question is: what is $y$ at $x=0$ for each function?

2. 2

   Read off G's initial value. $G$ is stated to pass through $(0,-1)$, and $x=0$ there, so

   $\text{initial value of }G=-1.$

   Its slope $\tfrac13$ is not needed for this comparison — it is a distractor.

3. 3

   Find H's slope from its two points.

   $m_H=\frac{4-3}{10-8}=\frac{1}{2}.$

4. 4

   Trace H back to x = 0. Starting from $(8,3)$ and moving $8$ units left, $y$ decreases by $8\times\tfrac12=4$:

   $\text{initial value of }H=3-4=-1.$

   Equivalently, $y=\tfrac12 x+b$ with $3=\tfrac12(8)+b$, so $b=-1$.

5. 5

   Compare and answer. Both initial values equal $-1$, so they are equal, not greater. Kelly is incorrect.

6. 6

   Verify with H's second point. The equation $y=\tfrac12 x-1$ gives $y=\tfrac12(10)-1=4$ at $x=10$ ✓, matching the given point. So the intercept $-1$ is right, and the two functions cross the $y$-axis at exactly the same place while rising at different rates.