<ol>
<li><p>I think this is impossible. I can think of a ton of different solutions from what you’re giving me. (1,18) and (2,18) would work, as would (1, 4.5) and (2,9), as would infiniti more solutions. does the y-intercept have to be at 0? That would change things. </p></li>
<li><p>y=-3x+2? Is the question asking to reflect over the y-axis? Because if it is, then it helps to draw it out (or imagine it). The y-intercept would remain the same, because a point on the y-axis reflected over the y-axis remains the same point. The slope would become negative, because that makes sense. Your y-values remain the same, hence changes in y remain the same, except the change in x to get your corresponding y change would be reveresed (confusing, ask me if you want me to elaborate).</p></li>
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<ol>
<li>If a linear function passes through the points (1, a) , (2, b) and (4, 18), what is the value of 3/2b-a</li>
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<p>3/(2b-a) right?</p>
<p>Well using the definition of slope and those points, I found that 6 = b - 2a. a = 1 doesn’t work, but a = 0 does. Then we get (1, 0) (2, 6) and (4, 18)</p>
<p>So 3/(2b-a) = 3/(2x6 - 0)
= 3/12
= 1/4</p>
<ol>
<li>Reflection in the y axis means that
(x, y) becomes (-x, y)</li>
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<p>y = 3x + 2
with the reflection…
y = 3(-x) + 2
y = -3x + 2</p>