Certainly! Let's match the attributes of each parabola to their corresponding functions.
Given the attributes, we have:
For the function [tex]\( f(x) = -(x-1)^2 + 4 \)[/tex]:
- Vertex: [tex]\((1, 4)\)[/tex]
- Focus: [tex]\(\left(1, 3 \frac{3}{4}\right)\)[/tex]
- Directrix: [tex]\( y = 4 \frac{1}{4} \)[/tex]
For the function [tex]\( f(x) = 2(x+1)^2 + 4 \)[/tex]:
- Vertex: [tex]\((-1, 4)\)[/tex]
- Focus: [tex]\(\left(-1, 4 \frac{1}{8}\right)\)[/tex]
- Directrix: [tex]\( y = 3 \frac{7}{8} \)[/tex]
Let's place these attributes in the table:
[tex]\[
\begin{tabular}{|l|l|}
\hline
$f(x)=-(x-1)^2+4$ & $f(x)=2(x+1)^2+4$ \\
\hline
Vertex: $(1, 4)$ & Vertex: $(-1, 4)$ \\
Focus: $\left(1, 3 \frac{3}{4}\right)$ & Focus: $\left(-1, 4 \frac{1}{8}\right)$ \\
Directrix: $y=4 \frac{1}{4}$ & Directrix: $y=3 \frac{7}{8}$ \\
\hline
\end{tabular}
\][/tex]
This correctly matches each attribute to the appropriate quadratic function.
