# Find the points on the curve y=2x^3+3x^2-12x+1 where the tangent is horizo

Find the points on the curve $$\displaystyle{y}={2}{x}^{{3}}+{3}{x}^{{2}}-{12}{x}+{1}$$ where the tangent is horizontal.

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nitruraviX
We are looking for a solution to this equation:
$$\displaystyle{y}'{\left({x}\right)}={0}$$
because the tangent is horizontal when the derivative is zero.
So,
$$\displaystyle{y}'{\left({x}\right)}={0}$$
$$\displaystyle{\left({2}{x}^{{3}}+{3}{x}^{{2}}-{12}{x}+{1}\right)}'={0}$$
$$\displaystyle{6}{x}^{{2}}+{6}{x}-{12}={0}$$
$$\displaystyle{x}^{{2}}+{x}-{2}={0}$$
Now we need to find a solution to this equation:
$$\displaystyle{x}^{{2}}+{x}-{2}={0}$$
This is a quadratic equation so the solution is given with this formula:
$$\displaystyle{x}_{{{1},{2}}}={\frac{{-{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}}$$
We have
$$\displaystyle{x}_{{{1},{2}}}={\frac{{-{1}\pm\sqrt{{{1}^{{2}}-{4}\cdot{1}\cdot{\left(-{2}\right)}}}}}{{{2}\cdot{1}}}}$$
$$\displaystyle{x}_{{{1},{2}}}={\frac{{-{1}\pm\sqrt{{9}}}}{{{2}}}}$$
$$\displaystyle{x}_{{{1}}}={\frac{{-{1}+{3}}}{{{2}}}}$$
$$\displaystyle{x}_{{1}}={1}$$
$$\displaystyle{x}_{{2}}={\frac{{-{1}-{3}}}{{{2}}}}$$
$$\displaystyle{x}_{{2}}={2}$$
So the given curve has horizontal tangent when x=1,-2, so the corresponding points are (1,-6) and (-2,21)
Result:
(1,-6), (-2,21)
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