find coordinate A from equation of the tangent and the gradient

Question

Q. The tangent to the curve $y=x^3-3x^2+x$ makes an angle of $\theta = 45^\circ$ with the positive direction of the $x$-axis Establish the coordinates of point $A$.

$$m=\tan\theta\\ m=\tan(45)\\ m=1$$

$$\frac{dy}{dx} = m\\ \frac{dy}{dx} =1\\ \frac{dy}{dx}= 3x^2-6x+1\\ 1=3x^2-6x+1\\ 0=3x^2-6x\\ 6x=3x^2\\ x=\frac{3x^2}{6}$$

Not sure what do to after this point?

Answer 1

Your problem is not with figuring out the tangent, its with algebra manipulation.

You have correctly figured out $3x^2−6x = 0$,

But why do this: $3x^2 = 6x$

, when you can factor to find solutions: $$3x^2−6x = 0 \to 3x(x−2) = 0$$

Hence, either, $x = 0, x - 2 = 0$, which gives the values for $x$: $x = {0, 2}$

Suggest reading... Factoring ... Factoring Quadratics ... Solving equations

Answer 2

Can be solved directly using differential calculus only.

( slope is the same as derivative at the required point.)

$$ y = 3x^3-3x^2+x$$ $$ y'(x)=9x^2-6x +1 =1\to x(3 x -2)=0$$

There are two points with slope =1.

$ x=(0, \frac{2}{3});$ The corresponding y values are $ y=(0,\frac{2}{9});$