## College Algebra (10th Edition)

For a square screen $\frac{X_{max}-X_{min}}{Y_{max}-Y_{min}}=1.6.$ Hence here $Y_{max}-Y_{min}=16/1.6=10$. Also we need $(4,8)$ to be contained, thus $Y_{min}\lt 8\lt Y_{max}$. Thus a possible solution is $Y_{min} = 0, Y_{max} = 10, and \ Y_{scl} = 1$.