math help please?

The answers are as follows:

y = (x² + 6)/(x + 6) ← this is a function, i.e. a curve

y(1) = (1 + 6)/(1 + 6)  = 7/7 = 1 → the representative curve of the function passes through A (1 ; 1)


The function f looks like (u/v), so the derivative looks like: [(u'.v) - (v'.u)]/v² → where:

u = x² + 6 → u' = 2x

v = x + 6 → v' = 1

f'(x) = [(u'.v) - (v'.u)]/v²

f'(x) = [2x.(x + 6) - (x² + 6)]/(x + 6)²

f'(x) = [2x² + 12x - x² - 6]/(x + 6)²

f'(x) = (x² + 12x - 6)/(x + 6)² ← this is the derivative

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

f'(1) = (1 + 12 - 6)/(1 + 6)² = 7/7² = 1/7 ← this is the slope of the tangent line to the curve at x = 1


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

The slope of the tangent line to the curve is: m = 1/7

The equation of the tangent line to the curve becomes: y = (1/7).x + b

The tangent line to the curve passes through A, so the coordinates of this point must verify the equation of the line.

y = (1/7).x + b

b = y - (1/7).x → you substitute x and y by the coordinates of the point A (1 ; 1)

b = 1 - [(1/7) * 1]

b = 6/7

→ The equation of the tangent line to the curve at x = 1 is: y = (1/7).x + (6/7)

y = (x + 6)/7

7y = x + 6

x - 7y = - 6

y'= (x² + 12x - 6)/(x + 6)²
a. 1/7
b. 1/7