Chapter 2.1, Problem 5QR

To determine

To calculate: The slope of the tangent to the parabola.

Answer to Problem 5QR

The slope of the tangent is $0$ .

Explanation of Solution

Given information:

The equation of parabola is $y=x^2+1$ at vertex $\left(0,0\right)$ .

Concept used:

The formula used is $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ and the vertex of a quadratic equation can be obtained by the formula $x=-\frac{b}{2a}$ and standard equation of parabola is $y=a{\left(x-h\right)}^2+k$ , where $(h,k)$ is vertex of the parabola.

Calculation:

The equation of parabola is $y=x^2+1$ .

Differentiate the function with respect to time $x$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left[x^2+1\right]\\ =2x+1\end{array}$

The slope of the tangent is the derivative.

  $m=2x$

On comparing the given equation, $y=x^2+1$ to the general equation of parabola, $y=a{\left(x-h\right)}^2+k$

It is obtained that $(h,k)=(0,1)$

Hence, the slope of the tangent at vertex $(h,k)=(0,1)$ is:

  $\begin{array}{c}m=2x\\ m=2\times 0\\ m=0\end{array}$

Conclusion:

The slope of tangent at vertex is $0$ .