harrylallamrita harrylallamrita
  • 10-12-2021
  • Mathematics
contestada

Find the values of the constant m for which the line y = mx is a tangent to the curve y=2x^2 - 4x + 8​

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Аноним Аноним
  • 10-12-2021

Answer:

m is 4 or -12.

Step-by-step explanation:

y = 2x^2 - 4x + 8.

To find the slope of the tangent at the point (x, y) we find the derivative.

y' = 4x - 4.

So m = 4x - 4.

Now our straight line is y = mx and m = 4x - 4 so

When y = mx meets the curve

mx = 2x^2 - 4x + 8

x(4x - 4) = 2x^2 - 4x + 8

4x^2 - 4x = 2x^2 - 4x + 8

2x^2 = 8

x^2 = 4

x = ±2.

So the line meets the curve and is tangent to it at the points where x = 2 and x = -2.

So the value of m   4(2) - 4 = 4

or 4(-2) - 4 = -12.

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