At what point on the parabola 𝑦=3𝑥2+2𝑥 is the tangent line parallel to the line 𝑦=10𝑥−2? Bold text start[6 marks]
Question
Solution
To find the point on the parabola where the tangent line is parallel to the line y=10x-2, we need to find where the derivative of the parabola is equal to the slope of the line.
Step 1: Find the derivative of the parabola The derivative of y=3x^2+2x is y'=6x+2.
Step 2: Set the derivative equal to the slope of the line The slope of the line y=10x-2 is 10. So we set the derivative equal to 10 and solve for x: 6x+2=10 6x=8 x=8/6=4/3
Step 3: Substitute x into the equation of the parabola to find y y=3*(4/3)^2+2*(4/3)=4+8/3=20/3
So the point on the parabola where the tangent line is parallel to the line y=10x-2 is (4/3, 20/3).
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