[tex]\begin{gathered} y=\frac{1}{4}(x-5)^2-3 \\ (x-5)^2=x^2+2(x)(-5)+(-5)^2 \\ (x-5)^2=x^2-10x+25 \\ Hence \\ y=\frac{1}{4}(x^2-10x+25)^{}-3 \\ y=\frac{1}{4}x^2-\frac{10x}{4}+\frac{25}{4}^{}-3 \\ y=\frac{1}{4}x^2-\frac{10x}{4}+\frac{13}{4}^{} \\ if\text{ a>0, it opens up, in this case a=}\frac{1}{4},\text{ and }\frac{1}{4}>0,\text{ therefore the equation opens upward} \\ \\ \text{Questions 2} \\ y=2(x-5)^2-3 \\ y=2(x^2-10x+25)-3 \\ y=2x^2-20x+50-3 \\ y=2x^2-20x+47 \\ \frac{1}{4}<2,\text{ then} \\ \text{The graph of the equation is wider} \end{gathered}[/tex]
