Question

How do you write \[y=3{{\left( x-1 \right)}^{2}}+5\] in standard form?

Answer 1

Hint: To write a given equation in its standard form we have to take all terms to one side of the equation, leaving only zero to the other side. We can do this by subtracting the terms present on any of the sides of the equation from both sides, till we get zero on that side. We will do the same for the given equation.

Complete step by step answer:
We are given the equation \[y=3{{\left( x-1 \right)}^{2}}+5\]. To write the expression to its standard form we have to take all terms to one side of the equation. We can do this for the given equation as follows
\[y=3{{\left( x-1 \right)}^{2}}+5\]
Subtracting \[y\] from both sides of the equation, we get
\[\Rightarrow y-y=3{{\left( x-1 \right)}^{2}}+5-y\]
\[\Rightarrow 0=3{{\left( x-1 \right)}^{2}}+5-y\]
Flipping the above equation, we get
\[\Rightarrow 3{{\left( x-1 \right)}^{2}}+5-y=0\]
We know that the expansion \[{{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\], using this expansion in the left-hand side of the above equation, it can be expressed as
\[\Rightarrow 3\left( {{x}^{2}}-2(x)(1)+{{1}^{2}} \right)-y+5=0\]
\[\Rightarrow 3\left( {{x}^{2}}-2x+1 \right)-y+5=0\]
Expanding the bracket on the left-hand side of the above equation, it can be expressed as
\[\Rightarrow 3{{x}^{2}}-3(2x)+3(1)-y+5=0\]
Simplifying the above equation, we get
\[\Rightarrow 3{{x}^{2}}-6x-y+8=0\]
Hence the above equation is the standard form of the given equation.

Note:
We can use the standard form of the equation to find various information from it. Let’s take an example of an equation of conics, the standard form of the equation of the conics is \[a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0\], we can find the type of the conic that the equation represents by expressing it in its standard form. by expressing in the standard form we can find the value of the coefficient of different terms.