Determine an equation of the line tangent to y=2x^2+3 that is parallel to the 8x-y+3=0.

Question

Determine an equation of the line tangent to y=2x^2+3 that is parallel to the 8x-y+3=0.

Answer 1

8x-y +3=0 --> y = 8x +3 = mx + b

A line parallel to y = 8x+3 will have the same slope, m=8. The slope is also the derivative of y=2x²+3

y' = 4x = 8 --> x = 2

y = 2(2)²+3 = 11. So the intersection point is (2,11)

To get the equation for the line going through that point

y-11 = 8(x-2) --> y = 8x-16+11 = 8x-5

y = 8x-5

Answer 2

The given line has a slope = 8.

You want to take the derivative of the quadratic and set it = 8.

y ' = 4x = 8 , x = 2

Lastly, find the point on the parabola (quadratic) at x = 2, and write the equation of the tangent line with slope = 8 containing that point (using pt-slope form of the line is easiest):

y = 2(2)^2 +3 = 11 , so (2 , 11) is the point of tangency and y - 11 = 8 (x - 2) is the equation of the tangent line.