the line y= 6x+1663 is tangent to graph of y= 2x^3 +6x^2-282x-1 at the point

Question

the line y= 6x+1663 is tangent to graph of y= 2x^3 +6x^2-282x-1 at the point

Answer 1

If the question is the point of tangency, set the derivative of the polynomial = slope of the line. Solve.

y = mx + b is the point slope form of a linear equation. The slope is m. The slope for y = 6x + 1663 has a slope 6.

y= 2x^3 +6x^2-282x-1 has a derivative 6x^2 + 12x -282.

6x^2 + 12x -282 = 6

x^2 + 2x - 47 = 1

x^2 + 2x - 48 = 0

(x+8)(x-6) = 0

x = -8 and x = 6

Plug those into the original equations to see which one makes the original equations equal. It will be -8. At 6 they have the same slope, but they don't intersect.