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Question: How do you write \(y = 5{x^2} - 3x + 2\) into vertex form?...

How do you write y=5x2−3x+2y = 5{x^2} - 3x + 2 into vertex form?

Explanation

Solution

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(x−h)2+ky = a{\left( {x - h} \right)^2} + k, where (h,k)\left( {h,k} \right) are the coordinates of the vertex and aa is a multiplier.
2. a2−2ab+b2=(a−b)2{a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}

Complete step by step solution:
The given equation is: y=5x2−3x+2y = 5{x^2} - 3x + 2
Now, we know that the general equation of a parabola in vertex form is:
y=a(x−h)2+ky = a{\left( {x - h} \right)^2} + k
First of all, we will take 5 common factors such that the coefficient of x2{x^2} becomes 1.
Thus, we get,
⇒y=5(x2−35x)+2\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:
⇒y=5[(x)2−2x(310)+(310)2−(310)2]+2\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 a2−2ab+b2=(a−b)2{a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}, we get,
⇒y=5[(x−310)2−9100]+2\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
⇒y=5(x−310)2−920+2\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,
⇒y=5(x−310)2+−9+4020\Rightarrow y = 5{\left( {x - \dfrac{3}{{10}}} \right)^2} + \dfrac{{ - 9 + 40}}{{20}}
⇒y=5(x−310)2+3120\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(x−h)2+ky = a{\left( {x - h} \right)^2} + k

Therefore, the given equation y=5x2−3x+2y = 5{x^2} - 3x + 2 in vertex form can be written as y=5(x−310)2+3120y = 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(x−h)2+ky = a{\left( {x - h} \right)^2} + k, where (h,k)\left( {h,k} \right) is the vertex. The "aa" in the vertex form is the same "aa" as in y=ax2+bx+cy = 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 (a±b)2=a2±2ab+b2{\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.