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Intives the slope and equation y=2x²-3x+1 at (-1,6)

User Ymn
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Final answer:

To find the slope and equation of a quadratic function at a specific point, we need to determine the derivative of the function and evaluate it at the given point. The slope at the point (-1,6) is -7, and the equation of the line that is tangent to the quadratic function at that point is y = -7x + 1.

Step-by-step explanation:

To find the slope and equation of a quadratic function at a specific point, we need to determine the derivative of the function and evaluate it at the given point.

  1. First, we find the derivative of the quadratic function y = 2x^2 - 3x + 1. The derivative of the function is given by dy/dx = 4x - 3.
  2. Next, we substitute the x-coordinate (-1) into the derivative to find the slope at that point: (4 * -1) - 3 = -4 - 3 = -7.
  3. Finally, we can plug the values of the point (-1,6) and the slope (-7) into the point-slope form of the equation of a line: y - y1 = m(x - x1). Substituting the values, we get y - 6 = -7(x - (-1)). Simplifying further gives the equation y = -7x + 1.

User Tyth
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