How do you find the equation of the tangent line to the graph of the function at a specific point given the equation y = 32 – 4x^2 + 5x + 6?

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To find the equation of the tangent line to the graph at the given point, we follow these steps:

  1. Identify the function: The function is given by y = 32 - 4x^2 + 5x + 6.
  2. Find the derivative: The derivative of the function, y', will give us the slope of the tangent line. We differentiate with respect to x:
    y' = -8x + 5
  3. Evaluate the derivative at the given point: We need to find the slope at the specific point where we want to determine the tangent line. Let’s assume the given point corresponds to a specific x value. For example, if we are given the point as (x_0, y_0), we would plug x_0 into the derivative to find the slope:
    m = -8(x_0) + 5
  4. Find the y value at x_0: Substitute x_0 back into the original function to find y_0:
    y_0 = 32 - 4(x_0)^2 + 5(x_0) + 6
  5. Write the equation of the tangent line: The equation of the tangent line can be expressed in point-slope form:
    y - y_0 = m(x - x_0)
    Replacing m, x_0, and y_0 with their respective values gives us the equation of the tangent line.

Here’s a quick example:

If we want to find the tangent line at the point where x = 1:
– First, compute the slope:
m = -8(1) + 5 = -3
– Then, find the corresponding y_0:
y_0 = 32 - 4(1)^2 + 5(1) + 6 = 39
– Finally, the equation of the tangent line is:
y - 39 = -3(x - 1) or y = -3(x - 1) + 39.

Thus, the equation of the tangent line at that point is y = -3x + 42.

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