What is the tangent of y=2x^2,(6,10) ?

The tangent of y=2x^2,(6,10) is y=4(6+sqrt(31))x-134-24sqrt(31) \lor y=4(6-sqrt(31))x+24sqrt(31)-134

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tangent of y=2x^2,(6,10)

tangent of

y = 2 x^{2}, (6, 10)

y = 4 (6 + 31) x - 134 - 2431 or y = 4 (6 - 31) x + 2431 - 134

Solution steps

Slope of the tangent line at

(6, 10) : m = \frac{2 x ^{2} - 10}{x - 6}

Find the slope of

y = 2 x^{2} : m = 4 x

Solve

4 x = \frac{2 x ^{2} - 10}{x - 6} : x = 6 + 31, x = 6 - 31

Find the tangent line at

x = 6 + 31 : y = 4 (6 + 31) x - 134 - 2431

Find the tangent line at

x = 6 - 31 : y = 4 (6 - 31) x + 2431 - 134

y = 4 (6 + 31) x - 134 - 2431 or y = 4 (6 - 31) x + 2431 - 134

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What is the tangent of y=2x^2,(6,10) ?
The tangent of y=2x^2,(6,10) is y=4(6+sqrt(31))x-134-24sqrt(31) \lor y=4(6-sqrt(31))x+24sqrt(31)-134