I have to find the tangent to $y=f(x)=x^3+2$ and should go through the origin.
I know that a line through the origin has the formula $y=ax$
And the equation for the tangent is $y=m(x-x_0)+y_0$
I take the derivative: $f'(x)=3x^2$
Hereafter I need help
On
Since $(x_0,y_0)$ is supposed to be on the graph, we have $y_0=x_0^3+2$. So the tangent equation becomes $$ y=m(x-x_0)+x_0^3+2 $$ Also, the tangent has the same derivative as the function at the point $x_0$, so $$m=f'(x_0)=3x^2_0$$Now the equation for the tangent is $$ y=3x^2_0(x-x_0) +x_0^3+2\\=3x^2_0x-3x_0^3+x_0^3+2\\ =3x_0^2x+ (2-2x_0^3) $$ Since the tangent must go through the origin, we must have $$ 2-2x_0^3=0 \implies x_0=1 $$
Suppose there is a point $(x_0, y_0)$ that is on the curve $y = x^3 + 2$. Then that means $y_0 = x_0^3 + 2$. Furthermore, the slope of the tangent line at this point $(x_0, y_0)$ is given by $f'(x_0) = 3x_0^2 = m$. Therefore, the equation of the tangent line through this point is given by $$y = m (x - x_0) + y_0 = 3x_0^2 (x - x_0) + x_0^3 + 2 = 3x_0^2 x - 2x_0^3 + 2.$$ Now we have to find the value of $x_0$ such that this line's $y$-intercept is zero; i.e. when $x = 0$ then $y = 0$. This requires $x_0$ to satisfy the condition $2x_0^3 = 2$. What value(s) of $x_0$ make this true? What is the resulting tangent line and the point on the curve that the tangent line passes through?