Question

How do you write $y = 3{\left( {x + 1} \right)^2} - 27$ in standard form?

Answer 1

Hint: Here, we are required to write the standard form of the given equation which is clearly in the vertex form. Thus, we will use the algebraic identity to expand the given bracket in the RHS and then, we will multiply all the terms inside the bracket by 3 as it is multiplying with the bracket. Then, solving the like terms further we will be able to find the required simplified standard form of the given equation.

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. ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$

Complete step-by-step answer:
The given equation is $y = 3{\left( {x + 1} \right)^2} - 27$
We can see that this equation is in the Vertex form: $y = a{\left( {x - h} \right)^2} + k$
Now, in order to convert the given equation in standard form, we will first of all, use the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ inside the bracket.
Thus, we get,
$y = 3\left( {{x^2} + 2x + 1} \right) - 27$
Now, multiplying each term inside the bracket by 3 as it is present outside the bracket.
Thus, we get,
$ \Rightarrow y = 3{x^2} + 6x + 3 - 27$
$ \Rightarrow y = 3{x^2} + 6x - 24$
This equation cannot be simplified further. Thus, this is in the standard form.

Therefore, we can write $y = 3{\left( {x + 1} \right)^2} - 27$ as $y = 3{x^2} + 6x - 24$ in the standard form.
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.