how do i find the coordinates?

I’ve differentiated to get 3x^2 - 6x + 1 but when i try to factorise for the x value i cannot do it, am i doing something wrong?

Update:

I have also done tan(45) to get the gradient of 1

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  • 1 month ago

    f(x) = x³ - 3x² + x ← this is a curve, i.e. a function

    f'(x) = 3x² - 6x + 1 ← this is the derivative

    …but the derivative is too the slope of the tangent line to the curve at x

    The tangent line to the curve makes an angle of 45 °.

    You know that: tan(45) = 1 → then you must solve for x the equation:

    f'(x) = 1

    3x² - 6x + 1 = 1

    3x² - 6x = 0

    3x.(x - 2) = 0

    x.(x - 2) = 0

    First case: x = 0 ← on the drawing, is not the case

    Second case: x = 2 ← this is the point the abscissa of the point A

    f(x) = x³ - 3x² + x → when: x = 2

    f(2) = 8 - 12 + 2 = - 2

    Point A (2 ; - 2)

    To go further, let's calculate the equation of the tangent line to the curve at x = 2

    The typical equation of a line is: y = mx + b → where m: slope and where y₀: y-intercept

    The slope of the tangent line is (1), so the equation of the tangent line becomes: y = x + y₀

    The tangent line passes through A (2 ; - 2), so these coordinates must verify the equation of the line.

    y = x + y₀

    y₀ = y - x → you substitute x and y by the coordinates of the point A (2 ; - 2)

    y₀ = - 2 - 2

    y₀ = - 4

    → The equation of the tangent line at A is: y = x - 4

  • 1 month ago

    y = x^3 - 3x^2 + x

    dy/dx = 3x^2 - 6x + 1

    1 = 3x^2 - 6x + 1

    0 = 3x^2 - 6x

    0 = 3x(x - 2)

    x = 0 or x = 2

    A is not on the y-axis so we can discard x = 0, leaving x = 2.

    y(2) = 2^3 - 3(2^2) + 2

    y(2) = 8 - 12 + 2 = -2

    A is at (2, -2).

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