Normals to Curves (Differentiation)

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SUMMARY

The discussion focuses on determining the point on the parabola defined by the equation y = 2x² - 7x - 15 where a normal line has a gradient of -1/2. The gradient of the tangent line at this point is calculated to be 2. By solving the derivative dy/dx = 4x - 7 = 2, the x-coordinate is found to be 2.25. Substituting this value back into the parabola equation yields the corresponding y-coordinate of -20.625, resulting in the point (2.25, -20.625).

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zebra1707
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Hi there
If possible can someone please review my understanding of this question.

Homework Statement



At what point on the Parabola y= 2x^2-7x-15 will a normal to the parabola have a gradient of -1/2

Homework Equations




The Attempt at a Solution



We are given the gradient of the normal as -1/2
Therefore gradient of the tangent = 2

dy/dx = 4x - 7 = 2
4x = 9
x = 2.25 (2 1/4)

plug into y = 2(2.25)^2 - 7(2.25) - 15
y = -20 5/8

Therefore the point on the parabola - where a normal to the parabola will have a gradient of
-1/2 = (2 1/4, -20 5/8).
 
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Hi zebra1707! :smile:

(try using the X2 tag just above the Reply box :wink:)
zebra1707 said:
At what point on the Parabola y= 2x^2-7x-15 will a normal to the parabola have a gradient of -1/2

We are given the gradient of the normal as -1/2
Therefore gradient of the tangent = 2

Therefore the point on the parabola - where a normal to the parabola will have a gradient of
-1/2 = (2 1/4, -20 5/8).

Looks good! :smile:
 

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