Final answer:
To find the equation of the tangent to the curve at the given point (0, 33), you need to find the slope of the curve at that point. The slope of a curve is equal to the derivative of the curve's equation with respect to t. Once you have the slope, you can use the point-slope form of a linear equation to find the equation of the tangent line.
Step-by-step explanation:
To find the equation of the tangent to the curve at the given point (0, 33), we need to find the slope of the curve at that point. The slope of a curve is equal to the derivative of the curve's equation with respect to t. So, we need to find the derivatives of x and y with respect to t, and then evaluate them at t = 0. We then use the point-slope form of a linear equation to find the equation of the tangent line.
Given that x = t^2 - 4t and y = t^2 + 4t + 1, we can find the derivatives as follows:
x' = 2t - 4
y' = 2t + 4
Substituting t = 0 into the derivatives, we get x' = -4 and y' = 4. The slope of the tangent line is therefore m = y' / x' = 4 / -4 = -1.
Using the point-slope form, we have y - y1 = m(x - x1), where (x1, y1) is the given point (0, 33). Substituting the values, we get y - 33 = -1(x - 0), which simplifies to y - 33 = -x. Rearranging the equation, we have y = -x + 33. Therefore, the equation of the tangent to the curve at the point (0, 33) is y = -x + 33.