Question

How do you write $y = 3{x^2} + x - 2$ in vertex form?

Answer 1

Hint: The given equation in the question is in the quadratic form of $y = a{x^2} + bx + c$ , thus we have to change the quadratic form into the vertex form. In order to do so, first we have to find out the coordinates of the vertex and then just substitute the values in the vertex formula. After doing some simplification we get the required answer.

Formula used: $y = a{(x - h)^2} + k$ , $y = a{x^2} + bx + c$, $ - \dfrac{b}{{2a}}$

Complete step-by-step solution:
It is given equation is: $y = 3{x^2} + x - 2$,
Here the given equation is in the form of the quadratic equation: $y = a{x^2} + bx + c$,
Thus we can find the values of the variable $a{\text{ and }}b$
Comparing both the equations, we get:
$a = 3$
$b = 1$
$c = - 2$
Now we need to find the coordinates of the vertex which is given as $(h,k)$ where $h$ represents the x coordinate and $k$ represents the y coordinates
The formula to find $h$ coordinate of the vertex is given as: $ - \dfrac{b}{{2a}}$
Substituting the values of $b$ and $a$, we get:
$h$=$ - \dfrac{b}{{2a}}$=$ - \dfrac{1}{{2 \times 3}} = - \dfrac{1}{6}$
Now, in order to find the value of $k$, we substitute the known values in $y = 3{x^2} + x - 2$
Here $x$is represented by $h$ coordinate and $y$ is represented by $k$, thus:
$k = 3{h^2} + h - 2$
On substituting the values of $h$, we get:
$\Rightarrow$$k = 3{\left( { - \dfrac{1}{6}} \right)^2} + \left( { - \dfrac{1}{6}} \right) - 2$
On further simplifying, we get:
$\Rightarrow$$k = 3\left( {\dfrac{1}{{36}}} \right) - \dfrac{1}{6} - 2$
On further simplifying we get:
$\Rightarrow$$k = \dfrac{1}{{12}} - \dfrac{1}{6} - 2$
Now, we take a common denominator and add all the terms:
$ \Rightarrow k = \dfrac{{1 - 2 - 12}}{{12}}$
Thus, we get:
$ \Rightarrow k = - \dfrac{{25}}{{12}}$
Therefore, $h = - \dfrac{1}{6}{\text{ and }}k = - \dfrac{{25}}{{12}}$
Putting these values in the vertex form equation $y = a{(x - h)^2} + k$ , we get:
\[y = 3{\left( {x - \left( { - \dfrac{1}{6}} \right)} \right)^2} - \dfrac{{25}}{{12}}\], where $a = 3$ , $h = - \dfrac{1}{6}$ and $k = - \dfrac{{25}}{{12}}$
On further simplifying, we get:
$y = 3{\left( {x + \dfrac{1}{6}} \right)^2} - \dfrac{{25}}{{12}}$
Hence we get the required answer.

The given equation in vertex form is $y = 3{\left( {x + \dfrac{1}{6}} \right)^2} - \dfrac{{25}}{{12}}$

Note: Alternate method to solve this equation: using the complete square method.
Here the given equation is: $y = 3{x^2} + x - 2$. This equation is in the form of the general quadratic equation $y = a{x^2} + bx + c$.
In order to solve this quadratic equation using the perfect square method, first we need to ensure that the coefficient of the variable ${x^2}$ becomes $1$. Given equation:
$y = 3{x^2} + x - 2$
Now let’s divide the numbers with variable $x$ with 3, in order to remove the coefficient of ${x^2}$, we get:
$y = 3\left( {{x^2} + \dfrac{1}{3}x} \right) - 2$
In order to make a perfect square, we need to multiply the coefficient of variable $x$ with $\dfrac{1}{2}$.
Thus: $y = 3\left( {{x^2} + \left( {\dfrac{1}{3} \times \dfrac{1}{2}} \right)x} \right) - 2$
On multiply the term and we get
$ \Rightarrow y = 3\left( {{x^2} + \left( {\dfrac{1}{6}} \right)x} \right) - 2$
Now in order to factorize the sum and balance it properly, we square the coefficient of the variable $x$ and add it within the brackets:
$ \Rightarrow y = 3\left( {{x^2} + \left( {\dfrac{1}{6}} \right)x + {{\left( {\dfrac{1}{6}} \right)}^2}} \right) - 2$
To balance out the $3{\left( {\dfrac{1}{6}} \right)^2}$ , we subtract the same amount from the right hand side. Thus the equation becomes:
$y = 3\left( {{x^2} + \left( {\dfrac{1}{6}} \right)x + {{\left( {\dfrac{1}{6}} \right)}^2}} \right) - 2 - 3{\left( {\dfrac{1}{6}} \right)^2}$
Now we factorize the variables in the form of ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
Here: $\left( {{x^2} + \left( {\dfrac{1}{6}} \right)x + {{\left( {\dfrac{1}{6}} \right)}^2}} \right)$ ⟶ ${x^2}$ corresponds to ${a^2}$ , $\dfrac{1}{6}x$ corresponds to $2ab$ and ${b^2}$ corresponds to ${\left( {\dfrac{1}{6}} \right)^2}$
Thus: \[y = 3{\left( {x + \dfrac{1}{6}} \right)^2} - 2 - 3{\left( {\dfrac{1}{6}} \right)^2}\]
In order to simplify it further:
\[y = 3{\left( {x + \dfrac{1}{6}} \right)^2} - 2 - 3 \times \dfrac{1}{{36}}\]
\[ \Rightarrow y = 3{\left( {x + \dfrac{1}{6}} \right)^2} - 2 - \dfrac{3}{{36}}\]
In order to simplify further, we reduce the terms:
\[ \Rightarrow y = 3{\left( {x + \dfrac{1}{6}} \right)^2} - 2 - \dfrac{1}{{12}}\]
Now taking common denominators, we add the numerical terms:
\[ \Rightarrow y = 3{\left( {x + \dfrac{1}{6}} \right)^2} - \dfrac{{24 - 1}}{{12}}\]
\[ \Rightarrow y = 3{\left( {x + \dfrac{1}{6}} \right)^2} - \dfrac{{25}}{{12}}\]
Now the above equation cannot be reduced any further.
It is now, in the form of the vertex equation given as: $y = a{(x - h)^2} + k$