Chapter 3.1, Problem 56E

To determine

To find:

The equation of the tangent

Answer to Problem 56E

The normal line at $x=0$ intersect the parabola only once

Explanation of Solution

Given:

The normal line to the parabola

  $y=x-x^2$

Concept used:

The equation is in slope −intercept form, $y=mx+c$

An equation for the line through the point $\left(x_1,y_1\right)$ with slope m is

  $y-y_1=m\left(x-x_1\right)$

Calculation:

The function

  $y=x-x^2\mathrm{.......................}\left(1\right)$

The derivative of a function

  $\begin{array}{l}y=f\left(x\right)\\ y^{\prime }=f^{\prime }\left(x\right)=\frac{dy}{dx}=m\end{array}$

Differentiating the equation (1) with respect to $x$

  $\begin{array}{l}y=x-x^2\\ y^{\prime }=1-2x\mathrm{........................}\left(2\right)\end{array}$

The slope of the tangent

  $\begin{array}{l}m=1-2\left(0\right)\\ m=1\end{array}$

The normal line is perpendicular to the tangent

  $m_{normal}=-\frac{1}{m_{\tan gent}}=-1$

An equation for the line through the point $\left(x_1,y_1\right)$ with slope m is

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-1=-1\left(x-0\right)\\ y=1-x\end{array}$

The line intersect the parabolas

  $\begin{array}{l}y=1-x\\ y=x-x^2\end{array}$

Solve through substitution

  $\begin{array}{l}1-x=x-x^2\\ x^2-2x+1=0\\ \left(x-1\right)\left(x-1\right)=0\\ x=1\end{array}$

The normal line at $x=0$ intersect the parabola only once