Question

what is the initial y - value of the function that models the relationship between the points (6, - 12) and (4, - 4)?

Explanation:

Step1: Find the slope

The slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \( m=\frac{y_2 - y_1}{x_2 - x_1} \). For the points \((6, - 12)\) and \((4,-4)\), we have \( x_1 = 6,y_1=-12,x_2 = 4,y_2=-4 \). So \( m=\frac{-4-(-12)}{4 - 6}=\frac{-4 + 12}{-2}=\frac{8}{-2}=-4 \).

Step2: Use point - slope form to find the equation

The point - slope form of a line is \( y - y_1=m(x - x_1) \). Let's use the point \((4,-4)\). So \( y-(-4)=-4(x - 4) \), which simplifies to \( y + 4=-4x+16 \), and then \( y=-4x + 12 \).

Step3: Find the initial y - value (y - intercept)

The initial y - value is the y - intercept of the line. For the equation \( y=-4x + 12 \), when \( x = 0 \), \( y=12 \)? Wait, no, wait. Wait, let's check the calculation again. Wait, \( y-(-4)=-4(x - 4)\) gives \( y + 4=-4x+16\), so \( y=-4x + 16 - 4=-4x+12 \). Wait, but let's use the other point \((6,-12)\) to verify. If \( x = 6 \), \( y=-4\times6+12=-24 + 12=-12 \), which matches. Now, the initial y - value is when \( x = 0 \), so \( y=-4\times0+12 = 12 \)? Wait, no, that can't be. Wait, maybe I made a mistake in slope calculation. Wait, \( m=\frac{-12-(-4)}{6 - 4}=\frac{-12 + 4}{2}=\frac{-8}{2}=-4 \). Then using point \((6,-12)\), \( y-(-12)=-4(x - 6)\), \( y + 12=-4x+24 \), \( y=-4x+12 \). Wait, when \( x = 0 \), \( y = 12 \). But let's check with the two points. For \((4,-4)\): \( y=-4\times4+12=-16 + 12=-4 \), which matches. For \((6,-12)\): \( y=-4\times6+12=-24 + 12=-12 \), which matches. So the initial y - value (y - intercept) is 12. Wait, but the options are 12, - 4,4, - 12. So the initial y - value is 12.

Wait, no, wait a second. Wait, maybe the function is a linear function, and the initial y - value is the y - intercept. The equation is \( y=-4x + 12 \), so when \( x = 0 \), \( y = 12 \). So the initial y - value is 12.

Answer:

12