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sqdancefan
You know the tangent point, so all you need for the tangent line equation is the slope. That is given by the derivative at x=3.

The derivative is
  y' = 3x^2 -2
At x=3, this is y' = 3(3²) -2 = 25

Then your line in point-slope form is
  y -22 = 25(x -3)
Ver imagen sqdancefan
Lanuel

An equation of the tangent line to the curve at the given points is [tex]y = 25x - 53[/tex]

Given the following data;

  • Point on the x-axis = 3
  • Point on the y-axis = 22
  • [tex]y = x^3 - 2x + 1[/tex]

In this exercise, you're required to find an equation of the tangent line to the curve at the given points.

First of all, we would differentiate the value of y at point x = 3.

[tex]y = x^3 - 2x + 1\\\\y' = 3x^2 - 2[/tex]

Substituting the value of x, we have;

[tex]y' = 3(3)^2 - 2\\\\y' = 3(9) - 2\\\\y' = 27 - 2[/tex]

y' = 25

Therefore, the slope (m) of the tangent line equation is 25.

Next, we would write an equation of the tangent line to the curve at the given points by using the formula;

[tex]y - y_{1} = m(x - x_1)[/tex]

Substituting the values into the formula, we have;

[tex]y - 22 = 25(x - 3)\\\\y - 22 = 25x - 75\\\\y = 25x - 75 + 22\\\\y = 25x - 53[/tex]

In conclusion, an equation of the tangent line to the curve at the given points is [tex]y = 25x - 53[/tex]

Find more information: https://brainly.com/question/19933259

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