How do you determine equations for the tangents to the points on the curve $y = - x^{4} + 8 x^{2}$ $y=-x^4+8x^2$ such that the tangent lines are perpendicular to the line x=1?

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How do you determine equations for the tangents to the points on the curve $y = - x^{4} + 8 x^{2}$ $y=-x^4+8x^2$ such that the tangent lines are perpendicular to the line x=1?

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Answer 1 · 2015-09-26T16:11:12

$y = 0 \land y = 16$ $y=0 ^^ y=16$

Explanation:

If the tangent lines are perpendicular to $x = 1$ $x=1$ they are parallel to x-axis. Slope of the x-axis is equal to zero and every line parallel with x-axis has the same slope, so $k = 0$ $k=0$ .

$k=y'=-4x^3+16x=0$
$-4x(x^2-4)=0 <=> x=0 vv x^2-4=0$
$x=0 vv x=-2 vv x=+2$

$y(0)=0 => y=0$
$y(-2)=16 => y=16$
$y(+2)=16 => y=16$

So, tangents are: $y=0 ^^ y=16$

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How do you determine equations for the tangents to the points on the curve $y = - x^{4} + 8 x^{2}$ $y=-x^4+8x^2$ such that the tangent lines are perpendicular to the line x=1?

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How do you determine equations for the tangents to the points on the curve y=-x^4+8x^2y=−x4+8x2y=-x^4+8x^2 such that the tangent lines are perpendicular to the line x=1?

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How do you determine equations for the tangents to the points on the curve $y = - x^{4} + 8 x^{2}$ $y=-x^4+8x^2$ such that the tangent lines are perpendicular to the line x=1?