Chapter 3.1, Problem 5QR

To determine

The slope of the line which is a tangent to the parabola at its vertex.

Answer to Problem 5QR

The required slope of the parabola is 0.

Explanation of Solution

Given information:

The parabola: $y=x^2+1$

Formula used:

The vertex form of a parabola:

  $y={\left(x-h\right)}^2+k$

The given equation of the parabola is,

  $y=x^2+1$.......… (1)

It is known that the vertex form of a parabola is $y={\left(x-h\right)}^2+k$ ; where $\left(h,k\right)$ are the coordinates of the vertex.

Rewrite the equation (1) as:

  $y={\left(x-0\right)}^2+1$

So, the vertex of the above equation is $\left(0,1\right)$ .

Use the above slope formula to find the slope at x = 0.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{\left[{\left(0+h\right)}^2+1\right]-\left(0^2+1\right)}{h}=\underset{h\to 0}{\lim }\frac{h^2+1-1}{h}\\ =\underset{h\to 0}{\lim }\frac{h^2}{h}\\ =\underset{h\to 0}{\lim }\left(h\right)\\ =0\end{array}$

Hence, the required slope of the parabola is 0.