3-1-23
X
00:00
/
00:00
CC
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.