Write a linear equation given (-2, 7) and (6, -5)
1 answer:
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<span>Differentiate implicitly:
</span>
<span>
Solve for y
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<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>
</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>
<span>Plugging this into the formula for y above gives the points
</span>
<span>which is where our tangent will be parallel to the x-axis.</span>
<span>
</span>