i'd just love someone to help me find R, please.?

Question

Curve C has the equation
y=(x-1)(x^2-4)

It cuts the x axis at the points (-2,0)(1,0)(2,0)

y=x+7 is an equation of the tangent to C at the point (-1,6)

The tangent to C at the point R is parallel to the tangent at the point (-1,6)

find the exact coordinates of R.


Impossible really.

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Answer 1

The tangent to the point R has the same slope as the tangent at the point (-1, 6). The tangent at (-1, 6) is given by y = x + 7, so the slope is 1.

So we want to find another point on the curve C such that the tangent at that point has slope equal to 1. The slope of the tangent at a point equals the slope of the curve at that point, and the slope of the curve is the derivative:

y = (x - 1)(x^2 - 4)
y' = (x - 1) * 2x + x^2 - 4
y' = 3x^2 - 2x - 4

So to find the coordinates where y' = 1, we solve the equation
3x^2 - 2x - 4 = 1
for x:

3x^2 - 2x - 5 = 0
3x^2 + 3x - 5x - 5 = 0
3x * (x + 1) - 5*(x + 1) = 0
(3x - 5)(x + 1) = 0

So x = 5/3 or x = -1. We already know the point (-1, 6) because it was given to us, so let's find the y-coordinate when x = 5/3:

(5/3 - 1) * ([5/3]^2 - 4) = -22/27

So the other point is
(5/3, -22/27)

Hope this helps.

Answer 2

Note that the slope of the tangent line at R is simply the derivative of C at the point R. Since it is parallel to the line y=x+7, its slope must be 1. So, we are looking for a point where the derivative of the curve C is equal to 1. Thus we have by the product rule:

1 = dy/dx = (x²-4) + (x-1)2x
1 = x² - 4 + 2x² - 2x
0 = 3x² - 2x - 5

Using the quadratic formula:

x = (2±√(4 + 4*3*5))/6
x = (2±√64)/6
x = (2±8)/6
x = -6/6 = -1 or x = 10/6 = 5/3

Now, we already know the tangent through the point at (-1, 6), R is the other point. So the x-coordinate of R is 5/3. To find the y-coordinate, just plug 5/3 into the equation for C:

y = (5/3 - 1)((5/3)² - 4)
y = (2/3)(25/9 - 4)
y = (2/3)(-11/9)
y = -22/27

So R = (5/3, -22/27). And we are done.

Answer 3

Actually the math involved is not that bad, you just have to wade through a lot of verbiage. The bottom line is that you are asked to find another line that is parallel to the line y=x+7. This means you are asked to find another line with the same slope as the first line; and the slope of this line is 1 (the standard way to write lines is y=mx+b where m is the slope, so the slope here is 1).

And, from calculus, you know the slope of a tangent line at a point is equal to the derivative. So, we need to differentiate your function and find where the derivative has a value of 1.

So, expand the function:
f(x)=x^3-x^2-4x+4

Differentiate:
f'(x)=3x^2 -2x -4

Normally, you set this equal to zero to find max and min, but here we want to know when this equals 1:

f'(x)=3x^2 -2x -4=1 => 3x^2-2x-5=0

we solve the quadratic:

x=[2 +/- Sqrt[4+60]/6 or x = -1 and 5/3

This is comforting, since we are already told one of our points is at x=-1. So, the only other place where there is a tangent line to the curve with a slope = 1 occurs when x=5/3 and y =-0.8152