Find the points on the curve y == 2x3 + 3x212x + 2 where the tangent line is horizontal.(x, y) = (| –5(smaller x-value)(х, у)22(larger x-value)Enhanced FeedbackdyPlease try again. Find the x-value where Y is equal to 0. This is where the tangent of the graph of the function is horizontal. So, setdxO and solve for x.dxNeed Help?Read ItWatch It

Name: Find the points on the curve y == 2x3 + 3x212x + 2 where the tangent
Uploaded: 2024-03-20T06:46:42.000Z
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Description: Find the points on the curve y == 2x3 + 3x212x + 2 where the tangent line is horizontal.(x, y) = (| –5(smaller x-value)(х, у)22(larger x-value)Enhanced FeedbackdyPlease try again. Find the x-value where Y is equal to 0. This is where the tangent of

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Math

Calculus

Find the points on the curve y = 2x + 3x2 - 12x + 2 where the tangent line is horizontal. (х, у) - -5 |(smaller x-value) (x, y) = 22 | (larger x-value) Enhanced Feedback Please try again. Find the x-value where is equal to 0. This is where the tangent of the graph of the function is horizontal. So, set = 0 and solve for x. xp