Find the equations of the tangents to the curve y= x(x-1)(x+2) at the points where the curve cuts the x axis

First of all, we compute the points of interest, i.e. the points where the curve cuts the x axis: since the expression is already factored, we have

Which means that the roots are

Next, we can expand the function definition:

In this form, it is much easier to compute the derivative:

If we evaluate the derivative in the points of interest, we have

This means that we are looking for the equations of three lines, of which we know a point and the slope. The equation

is what we need. The three lines are:

 This is the tangent at x = -2

 This is the tangent at x = 0

 This is the tangent at x = 1

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !