if y=1+2 x+6 x2...?

Question

y=1+2 x+6 x2.

Find the slope of the tangent line to the parabola at the point where x=4.

Answer 1

The slope of the tangent line at x is the value of the derivative at x. Since y = 6x^2 + 2x + 1, y' = 12x + 2. Evaluated at x = 4, that gives a slope of 50. (Or is this not a calculus question?)

Answer 2

your equation would not incorporate x1 or x2. a million / (x - a million) - a million / (x - 2) = a million/6 6(x - 2) - 6(x - a million) = (x - 2)(x - a million) 6x - 12 - 6x + 6 = (x - 2)(x - a million) -6 = (x - 2)(x - a million) x² - 3x + 2 = -6 x² - 3x - 4 = 0 (x - 4)(x + a million) = 0 x = 4 or x = -a million I take it your x1 and x2 are the two ideas?

Answer 3

y = 6 x^2 + 2x + 1
y = 6 x^2 + 2x + 1/6 + 1 - 1/6
y = [ sqr(6) x + 1/sqr(6) ] ^2 + 5/6
y - 5/6 = [ sqr(6) x + 1/sqr(6) ] ^2 . . . parabola facing up
. . . . . . vertex = (- 1/6 , 5 / 6)

when x = 4 . . . y = 105
y ' = 12 x + 2 . . . . . . substitute x = 4
y ' = 12(4) + 2 = 50
slope = 50