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21 the graph of a linear function passes through the points $(-1, -\fra…

Question

21 the graph of a linear function passes through the points $(-1, -\frac{1}{4})$ and $(1, -\frac{3}{4})$.

which equation represents the function?

a $y = -\frac{1}{2}x - \frac{1}{4}$

b $y = \frac{1}{2}x + \frac{1}{4}$

c $y = -\frac{1}{4}x - \frac{1}{2}$

d $y = \frac{1}{4}x + \frac{1}{2}$

Explanation:

Step1: Find the slope (m)

The formula for slope between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Let \((x_1, y_1)=(-1, -\frac{1}{4})\) and \((x_2, y_2)=(1, -\frac{3}{4})\). Then \(m=\frac{-\frac{3}{4}-(-\frac{1}{4})}{1 - (-1)}=\frac{-\frac{3}{4}+\frac{1}{4}}{2}=\frac{-\frac{2}{4}}{2}=\frac{-\frac{1}{2}}{2}=-\frac{1}{4}\)? Wait, no, wait, let's recalculate. Wait, \(-\frac{3}{4}-(-\frac{1}{4}) = -\frac{3}{4}+\frac{1}{4}=-\frac{2}{4}=-\frac{1}{2}\). Then denominator is \(1 - (-1)=2\). So \(m=\frac{-\frac{1}{2}}{2}=-\frac{1}{4}\)? No, wait, no, \(-\frac{1}{2}\) divided by 2 is \(-\frac{1}{4}\)? Wait, no, \(\frac{-\frac{1}{2}}{2}=-\frac{1}{4}\)? Wait, no, \(-\frac{1}{2} \times \frac{1}{2}=-\frac{1}{4}\). Wait, but let's check the options. Wait, maybe I made a mistake. Wait, the two points are \((-1, -\frac{1}{4})\) and \((1, -\frac{3}{4})\). So \(y_2 - y_1 = -\frac{3}{4}-(-\frac{1}{4}) = -\frac{3}{4}+\frac{1}{4}=-\frac{2}{4}=-\frac{1}{2}\). \(x_2 - x_1 = 1 - (-1)=2\). So slope \(m = \frac{-\frac{1}{2}}{2}=-\frac{1}{4}\)? Wait, no, that's not right. Wait, maybe I mixed up the points. Wait, let's use the slope formula again. \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take \((x_1, y_1)=(1, -\frac{3}{4})\) and \((x_2, y_2)=(-1, -\frac{1}{4})\). Then \(y_2 - y_1 = -\frac{1}{4}-(-\frac{3}{4}) = -\frac{1}{4}+\frac{3}{4}=\frac{2}{4}=\frac{1}{2}\). \(x_2 - x_1 = -1 - 1=-2\). Then \(m=\frac{\frac{1}{2}}{-2}=-\frac{1}{4}\). Hmm, same result. Wait, but let's check the options. The options have slopes like \(-\frac{1}{2}\), \(\frac{1}{2}\), \(-\frac{1}{4}\), \(\frac{1}{4}\). Wait, maybe I made a mistake in the points. Wait, the first point is \((-1, -\frac{1}{4})\) and the second is \((1, -\frac{3}{4})\). Let's plot them mentally. From x=-1 to x=1, x increases by 2, y decreases by \(-\frac{3}{4}-(-\frac{1}{4}) = -\frac{2}{4}=-\frac{1}{2}\). So slope is rise over run, which is \(-\frac{1}{2}\) over 2, so \(-\frac{1}{4}\)? Wait, no, rise over run is (change in y)/(change in x) = \(-\frac{1}{2}\)/2 = \(-\frac{1}{4}\). But let's check the options. Option B is \(y=\frac{1}{2}x+\frac{1}{4}\), slope \(\frac{1}{2}\). Option A: slope \(-\frac{1}{2}\). Wait, maybe I miscalculated the slope. Wait, let's do it again. \(y_2 - y_1 = -\frac{3}{4}-(-\frac{1}{4}) = -\frac{3}{4}+\frac{1}{4}=-\frac{2}{4}=-\frac{1}{2}\). \(x_2 - x_1 = 1 - (-1)=2\). So slope \(m = \frac{-\frac{1}{2}}{2}=-\frac{1}{4}\)? No, that's \(-\frac{1}{4}\), but option C has slope \(-\frac{1}{4}\). Wait, but let's check the y-intercept. Let's use point-slope form. \(y - y_1 = m(x - x_1)\). Let's take point \((-1, -\frac{1}{4})\) and slope \(m = -\frac{1}{4}\). Then \(y - (-\frac{1}{4}) = -\frac{1}{4}(x - (-1))\). So \(y + \frac{1}{4} = -\frac{1}{4}(x + 1)\). Distribute: \(y + \frac{1}{4} = -\frac{1}{4}x - \frac{1}{4}\). Subtract \(\frac{1}{4}\): \(y = -\frac{1}{4}x - \frac{1}{4} - \frac{1}{4}\)? Wait, no, \(-\frac{1}{4} - \frac{1}{4}=-\frac{2}{4}=-\frac{1}{2}\). So \(y = -\frac{1}{4}x - \frac{1}{2}\), which is option C? Wait, but let's check with the other point. Let's take x=1, plug into option C: \(y = -\frac{1}{4}(1) - \frac{1}{2}=-\frac{1}{4}-\frac{2}{4}=-\frac{3}{4}\), which matches the second point (1, -\frac{3}{4}). And for x=-1: \(y = -\frac{1}{4}(-1) - \frac{1}{2}=\frac{1}{4}-\frac{2}{4}=-\frac{1}{4}\), which matches the first point (-1, -\frac{1}{4}). So that works. Wait, but earlier when I calculated the slope, I thought maybe I made a mistake, but it's correct. So the equation is \(y = -\frac{1}{4}x - \frac{1}{2}\), which is option C.

Wait, b…

Answer:

C. \( y = -\frac{1}{4}x - \frac{1}{2} \)