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How do you express \[y={{\left( x-2 \right)}^{2}}+4\].

Answer
VerifiedVerified
547.8k+ views
Hint: The standard form of a function is the form which has all the variable terms and constant terms. To express the given equation in its standard form, we need to shift all the variable and constant terms to one side of the equation leaving only zero to the other side. To solve given problem, we need to use the expansion of the algebraic term \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\].

Complete step by step answer:
We are asked to express the equation \[y={{\left( x-2 \right)}^{2}}+4\] in its standard form. To do this, we need to take all the terms to one side of the equation. Subtracting y from both sides of equation, we get
\[\begin{align}
  & \Rightarrow y-y={{\left( x-2 \right)}^{2}}+4-y \\
 & \Rightarrow 0={{\left( x-2 \right)}^{2}}+4-y \\
\end{align}\]
Flipping the above sides of equation, we get
\[\Rightarrow {{\left( x-2 \right)}^{2}}+4-y=0\]
The left-hand side of the equation has term \[{{\left( x-2 \right)}^{2}}\]. This is of the form \[{{(a-b)}^{2}}\], we know the expansion of this is \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]. Here, we have \[a=x\And b=2\]. Substituting the values of a and b, we get \[{{\left( x-2 \right)}^{2}}={{x}^{2}}+{{2}^{2}}-2(x)(2)\]. We know that the square of 2 is 4, substituting this in above expression and, simplifying this expression, it can be written as \[{{\left( x-2 \right)}^{2}}={{x}^{2}}+4-4x\]. Substituting this expansion in the standard form of the function, we get
 \[\begin{align}
  & \Rightarrow {{\left( x-2 \right)}^{2}}+4-y=0 \\
 & \Rightarrow {{x}^{2}}+4-4x+4-y=0 \\
\end{align}\]
Simplifying the above equation, we get
\[\Rightarrow {{x}^{2}}-4x-y+8=0\]

Note: Expressing in standard forms also includes simplifying the expression. Standard forms of equations are very useful while solving some questions. For example, in coordinate geometry while solving questions of conics. we need to express the equation in its standard form, and check whether it satisfies conditions for a particular conic or not.