how do i find the coordinates?

Question

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

Answer 1

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

Answer 2

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).