Find parametric equations of the line that is tangent to the parabola y = x^2 at the point (−2, 4).

dy/dx = 2x

so at (-2,4) slope is -4

y-4 = -4(x+2)
y-4 = -4x - 8 = k

-4x-8 = k
-4x = k + 8

x = -2 - k/4
y = 4 + k

Parametric equations for linear function ??
Why not just
y = -4x -4 ??

Well, tangents and parabolas can be quite touchy. But fear not! I'll give it a shot with some clownish parametric equations.

Let's start by finding the derivative of the parabola y = x^2. Remember, the derivative gives us the slope of the tangent line at any given point.

The derivative of y = x^2 is dy/dx = 2x.

Now, let's find the slope at the point (-2, 4). We substitute x = -2 into the derivative: dy/dx = 2(-2) = -4. So the slope of the tangent at (-2, 4) is -4.

To find parametric equations, we need a starting point on the tangent line. We can choose (-2, 4) as our starting point.

Now, let's find the equation of the tangent line.

Using the point-slope form, we have y - 4 = -4(x + 2). Simplifying, we get y = -4x - 4.

To make it a bit more clownish, let's put it in parametric form!

Let t be the parameter that varies along the tangent line. Our x-coordinate becomes x = t, and our y-coordinate becomes y = -4t - 4.

So, the parametric equations of the tangent line to the parabola y = x^2 at the point (-2, 4) are:

x = t
y = -4t - 4

Hope that puts a clownish spin on the math for you!

To find the parametric equations of the line tangent to the parabola at the point (-2, 4), we need to find the slope of the tangent line at that point.

The derivative of the function y = x^2 will give us the slope of the tangent line at any given point. Differentiating y = x^2 with respect to x, we get:

dy/dx = 2x

Now we can substitute the x-coordinate of the given point (-2, 4) into the derivative to find the slope of the tangent line at that point:

m = dy/dx = 2(-2) = -4

So, the slope of the tangent line is -4.

Now we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is given by:

y - y1 = m(x - x1)

where (x1, y1) are the coordinates of the given point on the line, and m is the slope of the line.

Using the point (-2, 4) and the slope -4, we can write the equation of the tangent line as:

y - 4 = -4(x - (-2))

Simplifying this equation, we get:

y - 4 = -4(x + 2)

y - 4 = -4x - 8

y = -4x - 4

So, the equation of the tangent line is y = -4x - 4. To express this line as parametric equations, we can assign a parameter t to the x-coordinate:

x = t

Then, we can express y in terms of t by substituting t into the equation of the line:

y = -4t - 4

Therefore, the parametric equations of the line tangent to the parabola y = x^2 at the point (-2, 4) are:

x = t
y = -4t - 4

To find the parametric equations of the line that is tangent to the parabola y = x^2 at the point (-2, 4), we need to find the slope of the tangent line at that point and then use it to determine the parametric equations.

Step 1: Find the derivative of the equation y = x^2.
The derivative of y = x^2 is given by dy/dx = 2x.

Step 2: Evaluate the derivative at the given point (-2, 4).
Substituting x = -2 into the derivative, we have dy/dx = 2(-2) = -4.
This means that the slope of the tangent line at (-2, 4) is -4.

Step 3: Determine the parametric equations using the point-slope form.
The point-slope equation of a line is given by y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.

Substituting the point (-2, 4) and the slope -4 into the point-slope equation, we get:
y - 4 = -4(x - (-2))
y - 4 = -4(x + 2)
y - 4 = -4x - 8
y = -4x - 4

So, the parametric equations of the line tangent to the parabola y = x^2 at the point (-2, 4) are:
x = t
y = -4t - 4, where t is the parameter.