<p>I tend to miss these questions so I need to get this straight</p>
<p>When the question asks to flip an equation (say y = 3x + 1) over the x-axis, is this what I am supposed to do:</p>
<li>Change slope to opposite (negative to positive or positive to negative)</li>
<li>Change y-intercept to opposite (negative to positive or positive to negative)</li>
</ol>
<p>Which would mean the answer is y = -3x -1</p>
<p>And for the y-axis</p>
<li>Change slope to opposite (negative to positive or positive to negative)</li>
</ol>
<p>Which would mean the answer is y = -3x + 1</p>
<p>Can someone confirm?</p>
<p>If you have the function y = 3x + 1 ,its reflection is y = -3x-1 .It makes sense when you imagine it :)</p>
<p>Yes, that is correct.</p>
<p>"If you have the function y = 3x + 1 ,its reflection is y = -3x-1 .It makes sense when you imagine it " - That’s correct for reflecting over the x-axis only right?</p>
<p>Yeah, Y-Reflection would just be opposite reciprocal slope with the same y-intercept. I think…</p>
<p>@Sasquatch219
Do u mean opposite slope? (not opposite reciprocal)
For ex. if slope is 5
Opposite = -5
Opposite Reciprocal = -1/5</p>
<p>For x reflection, multiplying the equation by -1 does the trick. </p>
<p>Y reflection is not opposite reciprocal slope. I don’t think there’s any real “trick for them.” You just have to look and think about it.</p>
<p>@An0maly</p>
<p>Are you sure it’s not just multiplying the slope by -1 and keeping the rest (y-intercept) the same?</p>
<p>Yes. Consider the case y = x^2. The reflection about the y-axis is also y = x^2, not y = -x^2</p>
<p>^ok but for “normal” y = mx + b equations multiplying the slope by -1 and keeping the rest (y-intercept) the same should work</p>
<p>if you’re having trouble with the concept, just sketch a graph. it should make it easier once you can see what they want</p>