Answer:
-9-(-3)/0-(-3) = -3/3 = -1
y+9= -1(x-0) <u><em>Point Slope form</em></u>
y+9<u>-9</u>= -1x<u>-9</u>
y= -x-9 <em><u>Slope Intercept form</u></em>
y<u>+x+9</u> = -x-9 <u>+x+9</u>
y+x+9= 0 <u><em>General form</em></u>
Check the picture below.
is it even? well, even functions use the y-axis as a mirror, so a pre-image on the right-side, will be a mirror of the image on the left-side, but in this case it isn't so, if you put a mirror right on the y-axis, the left-side will look a bit different.
does it have a zero at x = 0? well, just look at the graph, is the line touching the x-axis at 0? nope.
does it have an asymptote at 0? well, surely you can see it right there.
<span>Differentiate implicitly:
</span>

<span>
Solve for y
</span>

<span>When the tangent is parallel to the x-axis we have y'=0, so we must solve
</span>

<span>To find the actual value of x we plug this expression for y into the original equation
</span>
![x^3-3x^3+27x^6=0 \\ \\x^3(27x^3-2)=0\implies x=\{0,{\sqrt[3]2\over3}\}](https://tex.z-dn.net/?f=x%5E3-3x%5E3%2B27x%5E6%3D0%0A%5C%5C%0A%5C%5Cx%5E3%2827x%5E3-2%29%3D0%5Cimplies%20x%3D%5C%7B0%2C%7B%5Csqrt%5B3%5D2%5Cover3%7D%5C%7D)
<span>Plugging this into the formula for y above gives the points
</span>
![(0,0)\text{ and }({\sqrt[3]2\over3},{\sqrt[3]4\over3})](https://tex.z-dn.net/?f=%280%2C0%29%5Ctext%7B%20and%20%7D%28%7B%5Csqrt%5B3%5D2%5Cover3%7D%2C%7B%5Csqrt%5B3%5D4%5Cover3%7D%29)
<span>which is where our tangent will be parallel to the x-axis.</span>
<span>
</span>
Step-by-step explanation:
-4 + 14 - 24 = -9t
-24 = -9t
divide both sides by -9
t = 14/9
Answer:
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Step-by-step explanation: