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3-1-23
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    To find the horizontal tangent, we're gonna use the limit as H approaches 0 of six X + H ^2 plus 4 * X + H -, 8 minus six X ^2 + 4 X -8 all over H. So that's going to simplify to the limit. As H approaches 0 of 6 * X ^2 + 2 XH plus H ^2 plus 4X plus 4H minus 8 -, 6 X squared minus four X + 8 all over H, That's going to turn into six X ^2 + 12 XH plus six H ^2 plus 4X plus 4H minus 8 -, 6 X squared minus four X + 8 all over H. So the limit as H goes to 0, the 6X squares are going to cancel, the power +8 and -8 are going to cancel. The -4 X and positive 4X are going to cancel. And if I divide an H out of each term, I get left a 12X plus six H + 4. So putting H goes to zero. There we get twelve X + 4. Now we want a horizontal tangent. If it's a horizontal tangent, that means that our slope has to equal 0 at a horizontal line. So if 0 equals twelve X + 4, we subtract the four and divide by 12 and we get X equal -4 twelfths, which reduces to negative 1/3. To find the Y value, we go back to the original and we stick in y = 6 * -1 third squared plus four times negative 1/3 - 8, so y = 6 * 1/9 - 4 thirds -8. Getting a common denominator here, we'd end up with six ninths is really 2/3 -, 4 thirds -24 thirds, so -26 thirds. So the point that we're looking for is going to be the point X is negative 1/3 and Y is -26 thirds.