A calculus problem by Lenedel Bert Delfin

The first step towards finding the normal line of the tangent line is, of course, to find the tangent line. To do this you must first find the slope by taking the derivative of the function.

$\frac{\mathrm d}{\mathrm d x} f(x) = \frac{\mathrm d}{\mathrm d x} (x^2 + 2x +1) = 2x + 2$

Next, since the point is at $(0, 0)$ , you substitute in $x=0$ .

$f'(0) = 2 * (0) + 2 = 2$

So the slope of the tangent line is $2$ . Substituting the slope and the point $(0, 0)$ into the point-slope form of a linear equation we get:

$(y - y_1) = m * (x - x_1) \to$ $(y - 0) = 2 * (x-0)$ )

The normal to a line is the negative reciprocal of the slope, so in this case the normal to a slope of $2$ is $\Large -\frac{1}{2}$ . Substituting that into the point slope equation we get:

$(y - 0) = -\frac{1}{2} * (x - 0) \to y = -\frac{1}{2} * x \to 2 * y = -x \to x + 2 * y = 0$