### Transformation of quadratic equation passing through the origin

Question Sample Titled 'Transformation of quadratic equation passing through the origin'

In the same rectangular coordinate system, the graph of ${y}={f{{\left({x}\right)}}}$ is drawn with dotted curve and the graph of ${y}={f{{\left({x}-{2}\right)}}}+{3}$ is drawn with solid curve. Which of the following may represent the graph of the coordinate plane?

A

${x}$${y}$
B

${x}$${y}$
C

${x}$${y}$
D

${x}$${y}$

 Note that the graph of ${y}={f{{\left({x}-{2}\right)}}}+{3}$ can be obtained  by translating the graph of ${y}$ $={f{{\left({x}\right)}}}$ ${2}$ units rightwards and ${3}$ units upwards.

 Assume that ${f{{\left({x}\right)}}}={a}{x}^{{2}}$, where ${a}$ is an arbitrary negative constant.  ${f{{\left({x}\right)}}}$ passes through the origin and opens downwards ∴  ${f{{\left({x}-{2}\right)}}}+{3}$ $={a}{\left({x}-{2}\right)}^{{2}}+{3}$ Therefore, the vertex of ${y}={f{{\left({x}-{2}\right)}}}+{3}$ is ${\left({2},{3}\right)}.$ ${y}={a}{\left({x}-{h}\right)}^{{2}}+{k}$ format ∴   Only the graph with vertex ${\left({2},{3}\right)}$ can be the possible solution.

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