1. Problem: Graph the line that contains the point $(-6, 1)$ and has a slope of $5$. 2. Recall the point-slope form of a line equation: $$y - y_1 = m(x - x_1)$$ where $(x_1, y_1)$ is a point on the line and $m$ is the slope. 3. Substitute $x_1 = -6$, $y_1 = 1$, and $m = 5$: $$y - 1 = 5(x - (-6))$$ which simplifies to $$y - 1 = 5(x + 6)$$. 4. Distribute the slope: $$y - 1 = 5x + 30$$. 5. Solve for $y$ to get slope-intercept form: $$y = 5x + 31$$. 6. This is the equation of the line to graph. It is not horizontal (not $y=4$). The original graph shown with $y=4$ is incorrect for this line. 7. To graph, plot the point $(-6,1)$. 8. Use the slope $5$ which means rise over run is $5/1$, so from $(-6,1)$ move $1$ unit right and $5$ units up to $( -5, 6 )$ and plot. 9. Draw the straight line through these points extending across the grid. Final Answer: the line equation is $$y = 5x + 31$$. The corrected graph contains points $(-6,1)$ at the left and $( -5, 6 )$ steeply rising, crossing the y-axis at $y=31$.