What are the coordinates of the point on the graph of f(x)=(x+1)(x+2) at which the tangent is parallel to the line with equation 3x-y-1=0?

Question

tkhunny · Answer

Have you considered the first derivative?

amorfide · Answer

since it mentions paralell
the gradient of f(x) will be equal to the gradient of 3x-y-1=0 (re arrange this into y=mx+b format)

so you will need to differentiate f(x) then make it equal to the gradient of the linear equation

then solve for x

then you can work out y and you have your coordinates

anonymous · Answer

yes I first got f(x)=x^2+3x+2 and then the derivative f'(x)=2x+3

anonymous · Answer

ok I will try that

anonymous · Answer

Ok so I had the derivative f'(x)=2x+3 and then I had that equal to the derivative of the second equation which is y'=3. So 2x+3=3. Then i solved for x and for x=0. would that be correct so far?

tkhunny · Answer

f(x) = (x+1)(x+2) = x^2 + 3x + 2

f'(x) = 2x + 3 = Gradient of f(x) for any given value of x

3x - y - 1= 0 ==> y = 3x - 1

Gradient of Line (Slope) = 3

Solve: f(x) gradient = Line Gradient
Solve: 2x + 3 = 3 ==> x = 0 This is the desired x-coordinate.

f(0) = 2 This is the desire y-coordinate.

Answer: (0,2)

Try to be WAY more organized and WAY less haphazard.