To find the equation of the tangent line to the curve at [tex]\( t = -2 \)[/tex], let's follow the steps:
### Step 1: Finding [tex]\( x \)[/tex] and [tex]\( y \)[/tex] coordinates at [tex]\( t = -2 \)[/tex]
The parameterized equations for the curve are:
[tex]\[
x = t^2 - 2t
\][/tex]
[tex]\[
y = t^3 - 3t
\][/tex]
First, we substitute [tex]\( t = -2 \)[/tex] into these equations to find the coordinates [tex]\((x, y)\)[/tex] at that point.
For [tex]\( x \)[/tex]:
[tex]\[
x = (-2)^2 - 2(-2) = 4 + 4 = 8
\][/tex]
For [tex]\( y \)[/tex]:
[tex]\[
y = (-2)^3 - 3(-2) = -8 + 6 = -2
\][/tex]
So, the coordinates at [tex]\( t = -2 \)[/tex] are [tex]\((8, -2)\)[/tex].
### Step 2: Finding the derivatives [tex]\( \frac{dx}{dt} \)[/tex] and [tex]\( \frac{dy}{dt} \)[/tex]
Next, we find the first derivatives of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] with respect to [tex]\( t \)[/tex].
For [tex]\( \frac{dx}{dt} \)[/tex]:
[tex]\[
\frac{dx}{dt} = 2t - 2
\][/tex]
For [tex]\( \frac{dy}{dt} \)[/tex]:
[tex]\[
\frac{dy}{dt} = 3t^2 - 3
\][/tex]
### Step 3: Evaluating the derivatives at [tex]\( t = -2 \)[/tex]
We substitute [tex]\( t = -2 \)[/tex] into the derivatives to find their values at that point.
For [tex]\( \frac{dx}{dt} \)[/tex]:
[tex]\[
\frac{dx}{dt} \bigg|_{t=-2} = 2(-2) - 2 = -4 - 2 = -6
\][/tex]
For [tex]\( \frac{dy}{dt} \)[/tex]:
[tex]\[
\frac{dy}{dt} \bigg|_{t=-2} = 3(-2)^2 - 3 = 3(4) - 3 = 12 - 3 = 9
\][/tex]
### Step 4: Finding the slope of the tangent line
The slope [tex]\( m \)[/tex] of the tangent line is given by the ratio of the derivatives:
[tex]\[
m = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{9}{-6} = -1.5
\][/tex]
### Step 5: Writing the equation of the tangent line
The general form of the equation of the tangent line is:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
Where [tex]\((x_1, y_1)\)[/tex] is the point on the curve and [tex]\( m \)[/tex] is the slope we calculated.
Substitute [tex]\( m = -1.5 \)[/tex], [tex]\( x_1 = 8 \)[/tex], and [tex]\( y_1 = -2 \)[/tex]:
[tex]\[
y - (-2) = -1.5(x - 8)
\][/tex]
Simplify:
[tex]\[
y + 2 = -1.5(x - 8)
\][/tex]
Thus, the equation of the tangent line is:
[tex]\[
y = -1.5(x - 8) - 2
\][/tex]
This can be further simplified to:
[tex]\[
y = -1.5x + 12 - 2 = -1.5x + 10
\][/tex]
Or in the form we found initially:
[tex]\[
y = -1.5(x - 8) - 2
\][/tex]
This can also be left as is:
[tex]\[
y = -1.5 \cdot (x - 8) + (-2)
\][/tex]
Therefore, the equation of the line tangent to the curve at [tex]\( t = -2 \)[/tex] is:
[tex]\[
y = -1.5 \cdot (x - 8) + (-2)
\][/tex]
