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Find the values of k for which y = kx + 1 is a tangent to the curve y = 2x^2 + x +3

User Preben
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2 Answers

5 votes
k = 5 or k = 3
Hope it helps-,-
User AvnerSo
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2 votes

Final answer:

The values of k for which the line y = kx + 1 is a tangent to the curve y = 2x² + x + 3 are 1 + sqrt(8) and 1 - sqrt(8).

Step-by-step explanation:

To find the values of k for which the line y = kx + 1 is a tangent to the curve y = 2x² + x + 3, we need to find the point(s) of intersection between the line and the curve.

Let's equate the two equations and solve for x:

2x² + x + 3 = kx + 1

2x² + (1-k)x + 2 = 0

For the line to be tangent to the curve, this quadratic equation should have exactly one solution.

This means the discriminant (b² - 4ac) should be equal to 0.

Using the discriminant formula, (1 - k)² - 4(2)(2) = 0

Simplifying, we get k² - 2k - 7 = 0

Using the quadratic formula,

k = (-b ± sqrt(b² - 4ac))/(2a)

k = (2 ± sqrt((-2)² - 4(1)(-7)))/(2(1))

k = (2 ± sqrt(32))/(2)

k = 1 ± sqrt(8)

Therefore, the values of k for which the line is a tangent to the curve are 1 + sqrt(8) and 1 - sqrt(8).

User Jason Fingar
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7.8k points

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