Answer:
a)(0,k)
b)(0,-k)
Step-by-step explanation:
Finding the vertex: Using calculus
Given:
[tex] \begin{cases} y = a {x}^{2} + k \\ y = a {x}^{2} - k \end{cases}[/tex]
Taking the derivative of y's yields
[tex] \begin{cases} y '= \dfrac{d}{dx} a {x}^{2} + k \\ \\ y '= \dfrac{d}{dx} a {x}^{2} - k \end{cases} \\ \implies\begin{cases} y '= 2a x\\ \\ y '=2ax\end{cases} [/tex]
Now set y' to 0 and solve for x:
[tex] \implies\begin{cases} 2a x = 0\\ \\ 2ax = 0\end{cases} \\ \implies\begin{cases} x= 0\\ \\ x= 0\end{cases} [/tex]
plug in the value of x into the original equation thus
[tex] \begin{cases} y = a ({0}^{2}) + k \\ y = a ({0}^{2} )- k \end{cases}\\ \implies \begin{cases} y = k \\ y= - k \end{cases} [/tex]
Hence,
- a. vertex (0,k)
- b. vertex (0,-k)
