for f(x)= 1-x-x^2, a=0 find tangent slope of tangent line?

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f(x) = 1 - x - x² ← this is a curve

f(a) = 1 - a - a² → the curve passes through the point M [a ; (1 - a - a²)]


f'(x) = - 1 - 2x ← this is the derivative

…but the derivative is too the slope of the tangent line to the curve at x

f'(a) = - 1 - 2a ← this is the slope of the tangent line to the curve at: x = a


The typical equation of a line is: y = mx + y₀ → where m: slope and where y₀: y-intercept

The slope of the tangent line is (- 1 - 2a).

The equation of the tangent line becomes: y = (- 1 - 2a).x + y₀

The tangent line passes through M, so these coordinates must verify the equation of the tangent line.

y = (- 1 - 2a).x + y₀

y₀ = y - (- 1 - 2a).x → you substitute x and y by the coordinates of the point M [a ; (1 - a - a²)]

y₀ = (1 - a - a²) - (- 1 - 2a).a

y₀ = 1 - a - a² + a + 2a²

y₀ = a² + 1

The equation of the tangent line to the curve (at x = a) is:

y = (- 1 - 2a).x + (a² + 1)

y = - (2a + 1).x + (a² + 1)...Show more