Question

the table shows a function. is the function linear or nonlinear?
x | -10 | 6\frac{1}{2} | 9\frac{1}{2}
y | 2\frac{3}{4} | -5\frac{1}{2} | -7
linear
nonlinear

Explanation:

Step1: Find Δx (x differences)

First pair: $6\frac{1}{2} - (-10) = \frac{13}{2} + 10 = \frac{13}{2} + \frac{20}{2} = \frac{33}{2}$
Second pair: $9\frac{1}{2} - 6\frac{1}{2} = 3 = \frac{6}{2}$

Step2: Find Δy (y differences)

First pair: $-5\frac{1}{2} - 2\frac{3}{4} = -\frac{11}{2} - \frac{11}{4} = -\frac{22}{4} - \frac{11}{4} = -\frac{33}{4}$
Second pair: $-7 - (-5\frac{1}{2}) = -7 + \frac{11}{2} = -\frac{14}{2} + \frac{11}{2} = -\frac{3}{2}$

Step3: Check slope consistency

Slope 1: $\frac{\Delta y}{\Delta x} = \frac{-\frac{33}{4}}{\frac{33}{2}} = -\frac{33}{4} \times \frac{2}{33} = -\frac{1}{2}$
Slope 2: $\frac{\Delta y}{\Delta x} = \frac{-\frac{3}{2}}{\frac{6}{2}} = -\frac{3}{6} = -\frac{1}{2}$
*(Note: To confirm linearity, we can also verify with the linear equation. Using point $(-10, 2\frac{3}{4})$ and slope $-\frac{1}{2}$:
Equation: $y - \frac{11}{4} = -\frac{1}{2}(x + 10)$
Simplify: $y = -\frac{1}{2}x - 5 + \frac{11}{4} = -\frac{1}{2}x - \frac{20}{4} + \frac{11}{4} = -\frac{1}{2}x - \frac{9}{4}$
Test $x=6\frac{1}{2}=\frac{13}{2}$: $y = -\frac{1}{2}(\frac{13}{2}) - \frac{9}{4} = -\frac{13}{4} - \frac{9}{4} = -\frac{22}{4} = -5\frac{1}{2}$, which matches.
Test $x=9\frac{1}{2}=\frac{19}{2}$: $y = -\frac{1}{2}(\frac{19}{2}) - \frac{9}{4} = -\frac{19}{4} - \frac{9}{4} = -\frac{28}{4} = -7$, which matches.)*

Answer:

linear