Question

How do you convert $y=3{{\left( x-5 \right)}^{2}}+5$to standard form ?

Answer 1

Hint: For finding the standard form of the given equation $y=3{{\left( x-5 \right)}^{2}}+5$ , we will have to expand the given equation $y=3{{\left( x-5 \right)}^{2}}+5$ in which we will use the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ for ${{\left( x-5 \right)}^{2}}$ and will have to convert the given equation $y=3{{\left( x-5 \right)}^{2}}+5$ in the form of $y=a{{x}^{2}}+bx+c$, where $a$, $b$ and $c$ are constants.

Complete step by step answer:
Since, we have the question in the form of equation as:
$\Rightarrow y=3{{\left( x-5 \right)}^{2}}+5$
Now, we will expand the given equation with use of the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ as:
$\Rightarrow y=3\left[ {{\left( x \right)}^{2}}+{{\left( 5 \right)}^{2}}-2\times x\times 5 \right]+5$
Here, we will get ${{x}^{2}}$ and $25$ after square $x$ and $5$ respectively. So, we can write the above equation as:
$\Rightarrow y=3\left[ {{x}^{2}}+25-2\times x\times 5 \right]+5$
We will use multiplication method for $2\times x\times 5$ and will get $10x$ as:
$\Rightarrow y=3\left[ {{x}^{2}}+25-10x \right]+5$
Now, we will open the bracket to solve the above equation as:
$\Rightarrow y=3\times {{x}^{2}}+3\times 25-3\times 10x+5$
Here, we will get $3{{x}^{2}}$ , $75$ and $30x$ when we will complete the multiplication process of the above equation as:
$\Rightarrow y=3{{x}^{2}}+75-30x+5$
Since, $75$ and $5$ are equal like terms. So, we will add them as:
$\Rightarrow y=3{{x}^{2}}-30x+\left( 75+5 \right)$
$\Rightarrow y=3{{x}^{2}}-30x+80$
Above equation is in the standard form of the given equation $y=3{{\left( x-5 \right)}^{2}}+5$. We will compare this standard form of the given equation with the symbolic standard form $y=a{{x}^{2}}+bx+c$. So, we will get:
$a=3$ , $b=-30$ and $c=80$ .
Hence, the equation $y=3{{x}^{2}}-30x+80$ is the standard form of the given equation $y=3{{\left( x-5 \right)}^{2}}+5$ .

Note:
Since, we have the given equation $y=3{{\left( x-5 \right)}^{2}}+5$ from the question and the standard form of the given equation $y=3{{x}^{2}}-30x+80$ . So, we can write it as:
\[\Rightarrow y=3{{\left( x-5 \right)}^{2}}+5\Leftrightarrow y=3{{x}^{2}}-30x+80\]
Where, L.H.S. $\Rightarrow y=3{{\left( x-5 \right)}^{2}}+5$ and R.H.S. $\Rightarrow y=3{{x}^{2}}-30x+80$ .
Now, we will check it whether the both the linear equation are same or not in the following way:
Let us consider a value of $x$ . Here, we assume that $x$ is equal to $1$ as:
Now, we will put the value of $x$ in the equation \[y=3{{\left( x-5 \right)}^{2}}+5\Leftrightarrow y=3{{x}^{2}}-30x+80\] also:
\[\Rightarrow y=3{{\left( 1-5 \right)}^{2}}+5\Leftrightarrow y=3{{\left( 1 \right)}^{2}}-30\times 1+80\]
\[\Rightarrow y=3{{\left( -4 \right)}^{2}}+5\Leftrightarrow y=3\times 1-30+80\]
\[\Rightarrow y=3\times 16+5\Leftrightarrow y=3-30+80\]
\[\Rightarrow y=53\Leftrightarrow y=53\]
Since, we got$y=53$ with respect to $x=1$for both the equations that defines that L.H.S. is equal to R.H.S. Hence, the solution is correct.