Question

Write an equation in standard form of the parabola that has the same shape as the graph of f (x)=6x squared


`f(x) = 6 {x}^(2)`
, but with (2,5) as the vertex

Answer 1

`\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\ x=a(y-{{ k}})^2+{{ h}} \end{array} \qquad\qquad vertex~~(\stackrel{}{{{ h}}},\stackrel{}{{{ k}}})\\\\ -------------------------------\\\\ f(x)=6x^2\implies f(x)=6(x-\stackrel{h}{0})^2\stackrel{k}{+0} \\\\\\ now\qquad \begin{cases} h=2\\ k=5 \end{cases}\implies f(x)=6(x-2)^2+5 \\\\\\ f(x)=6(x^2-2x+2^2)+5\implies f(x)=6x^2-12x+24+5 \\\\\\ f(x)=6x^2-12x+29`