For the curve $y=x^2+3x+4$, find all the points at which the tangent passes through the origin.
$y=x^2+3x+4$ $\;\; \cdots$ (1)
Differentiating equation (1) w.r.t x gives
$\dfrac{dy}{dx}=2x+3$
The required tangent passes through the origin $\left(0,0\right)$
$\therefore$ $\;$ Equation of tangent is
$y-0 = \left(2x+3\right) \left(x-0\right)$
i.e. $y=2x^2 + 3x$ $\;\; \cdots$ (2)
Solving equations (1) and (2) simultaneously gives
$2x^2 + 3x = x^2 + 3x + 4$
i.e. $x^2 = 4$ $\implies$ $x = \pm 2$
Substituting the value of x in equation (1) gives
when $x=+2$, $y= \left(2\right)^2 + 3 \times 2 + 4 = 14$
when $x=-2$, $y = \left(-2\right)^2 + 3 \times \left(-2\right) + 4 = 2$
$\therefore$ $\;$ The required points are $\left(2,14\right)$ and $\left(-2,2\right)$