The problem was to find the equations of all the lines passing through the point,
(1,4), that are tangent to the curve,
y(x) = x3 - 10x2 + 6x - 2
as well as each point of tangency.
Step 1: Find the derivative of y(x).
y'(x) = 3x2 - 20x + 6
Step 2: Set up the equation to solve for the
x values of the points
of tangency. To constrain a line to pass through the point,
(1,4),
we use this equation for the line:
y - 4 = m(x - 1)
Assuming that
x in the above is the
x coordinate of a point
of tangency, we know that
m must, at that point, equal to the derivative,
y'(x). Hence we substitute
m with the expression for
y'(x).
y - 4 = (3x2 - 20x + 6)(x - 1)
Assuming again that the
x is the
x coordinate of a point of
tangency, we know that the
y value of line and the
y value of the
function,
y(x), must coincide at
x. Hence we substitute the
expression given in the problem for
y:
x3 - 10x2 + 6x - 2 - 4 = (3x2 - 20x + 6)(x - 1)
Step 3: Find all the solutions for
x. First multiply out the right
hand side of this thing.
x3 - 10x2 + 6x - 6 = 3x3 - 20x2 + 6x - 3x2 + 20x - 6
There's a few terms that cancel from both sides:
x3 - 10x2 = 3x3 - 20x2 - 3x2 + 20x
Now subtract the left from the right and gather like terms:
0 = 2x3 - 13x2 + 20x
And as cubics go, this one is pretty easy. Since there's no constant term, factor out an
x
0 = x(2x2 - 13x + 20)
So
x = 0 is certainly a solution. Now apply the
quadratic formula to the remaining factor.
_________
13 ± √169 - 160 5
x = = & 4
4 2
which gives you the remaining two solutions for
x. So the three solutions are
Step 4: Find the slopes of the lines. There are lines that satisfies the
conditions of the problem for each of the solutions above for
x. We find
them by plugging those solutions into the expression we got for
y'(x).
y'(x) = 3x2 - 20x + 6
So here we go finding the slopes:
m1 = 3x12 - 20x1 + 6 = 6
m2 = 3x22 - 20x2 + 6 =
|
75 100 101
- + 6 = - = -25.25
4 2 4
|
m3 = 3x32 - 20x3 + 6 = 48 - 80 + 6 = -26
Step 5: Find the intercepts for each of the solutions. Do this by solving
4 = m + b
for each of
m1,
m1, and
m3.
This will give you
b1,
b2, and
b3, from which you can complete the equations of each of the
three lines (
y = mx + b) that are solutions to this problem. I won't do this step out for you
because it's very easy.
Step 6: Find the points of tangency. This is just plugging the three
x's we got above back into the original y(x):
y(x) = x3 - 10x2 + 6x - 2
y1 = x13 - 10x12 + 6x1 - 2 = -2
y2 = x23 - 10x22 + 6x2 - 2 =
|
125 250 30
- + - 2 = -
8 4 2
|
271
8
|
y3 = x33 - 10x32 + 6x3 - 2 = 64 - 160 + 24 - 2 = -74
The points of tangency are, of course,
(x1,y1),
(x2,y2), and
(x3,y3).
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