Question

How do you write $y = 5{x^2} - 3x + 2$ into vertex form?

Answer 1

Hint: Here, we will use the completing the square method and simplify the given equation. Then we will compare the obtained equation to the general vertex form of a parabola. We will simplify it further to get the required answer. A quadratic equation is an equation that has the highest degree of 2 and has two solutions.

Formula Used:
We will use the following formulas:
1. Vertex form: $y = a{\left( {x - h} \right)^2} + k$, where $\left( {h,k} \right)$ are the coordinates of the vertex and $a$ is a multiplier.
2. ${a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}$

Complete step by step solution:
The given equation is: $y = 5{x^2} - 3x + 2$
Now, we know that the general equation of a parabola in vertex form is:
$y = a{\left( {x - h} \right)^2} + k$
First of all, we will take 5 common factors such that the coefficient of ${x^2}$ becomes 1.
Thus, we get,
$ \Rightarrow y = 5\left( {{x^2} - \dfrac{3}{5}x} \right) + 2$
Hence, using completing the square method and adding and subtracting the square of a constant, we can rewrite the given equation as:
$ \Rightarrow y = 5\left[ {{{\left( x \right)}^2} - 2x\left( {\dfrac{3}{{10}}} \right) + {{\left( {\dfrac{3}{{10}}} \right)}^2} - {{\left( {\dfrac{3}{{10}}} \right)}^2}} \right] + 2$
Thus, using the identity ${a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}$, we get,
$ \Rightarrow y = 5\left[ {{{\left( {x - \dfrac{3}{{10}}} \right)}^2} - \dfrac{9}{{100}}} \right] + 2$
Multiplying each term inside the bracket by 5, we get
$ \Rightarrow y = 5{\left( {x - \dfrac{3}{{10}}} \right)^2} - \dfrac{9}{{20}} + 2$
Taking the LCM of the denominators of the constants, we get,
$ \Rightarrow y = 5{\left( {x - \dfrac{3}{{10}}} \right)^2} + \dfrac{{ - 9 + 40}}{{20}}$
$ \Rightarrow y = 5{\left( {x - \dfrac{3}{{10}}} \right)^2} + \dfrac{{31}}{{20}}$
Thus, we can say that this equation is in the general form $y = a{\left( {x - h} \right)^2} + k$

Therefore, the given equation $y = 5{x^2} - 3x + 2$ in vertex form can be written as $y = 5{\left( {x - \dfrac{3}{{10}}} \right)^2} + \dfrac{{31}}{{20}}$
Hence, this is the required answer.

Note:
The vertex form of a quadratic is given by $y = a{\left( {x - h} \right)^2} + k$, where $\left( {h,k} \right)$ is the vertex. The "$a$" in the vertex form is the same "$a$" as in $y = a{x^2} + bx + c$ (i.e., both have exactly the same value). We convert a given equation to its vertex form by completing the square. To complete the square, we try to make the identity ${\left( {a \pm b} \right)^2} = {a^2} \pm 2ab + {b^2}$ by adding or subtracting the square of constants and hence, taking the constant on the RHS to find the required simplified equation.