Question

How do you write the standard form of the equation of the parabola that has the indicated vertex and whose graph passes through the given point: Vertex: $\left( {6,5} \right)$; point: $\left( {0, - 4} \right)$?

Answer 1

Here, we are required to write the standard form of the given equation of the parabola that has the indicated vertex $\left( {6,5} \right)$ and whose graph passes through the given point $\left( {0, - 4} \right)$. Thus, we will write the general vertex form of a parabola and then, we will substitute the vertex and after substituting the given point, we will be able to find the multiplier $a$which when substituted again in the first equation, will give us the vertex form which when simplified further, will give us the required standard form.

Formula Used:
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:
As we know, the general equation of a parabola in vertex form is:
$y = a{\left( {x - h} \right)^2} + k$
Where, $\left( {h,k} \right)$are the coordinates of the vertex and $a$is a multiplier.
Now, according to the question, we have,
Vertex: $\left( {6,5} \right)$
Therefore, substituting $\left( {h,k} \right) = \left( {6,5} \right)$we get,
$y = a{\left( {x - 6} \right)^2} + 5$………………………….$\left( 1 \right)$
Now, in order to find the point $a$, we will substitute the given point such that $\left( {x,y} \right) = \left( {0, - 4} \right)$, thus, we get,
$- 4 = a{\left( {0 - 6} \right)^2} + 5$
$\Rightarrow - 4 = 36a + 5$
Subtracting 5 from both sides,
$\Rightarrow - 9 = 36a$
Dividing both sides by 36,
$\Rightarrow a = - \dfrac{1}{4}$
Therefore, the value of $a$is $- \dfrac{1}{4}$
Hence, substituting this in $\left( 1 \right)$, we get,
$y = - \dfrac{1}{4}{\left( {x - 6} \right)^2} + 5$
Thus, the vertex form is $y = - \dfrac{1}{4}{\left( {x - 6} \right)^2} + 5$
Now, in order to convert this to standard form, i.e. $y = A{x^2} + Bx + C$, first of all, we will use the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$in the bracket in the RHS.
Thus, we get,
$y = - \dfrac{1}{4}\left( {{x^2} - 12x + 36} \right) + 5$
Now, opening the bracket and multiplying each term by $- \dfrac{1}{4}$, we get,
$\Rightarrow y = - \dfrac{1}{4}{x^2} + 3x - 9 + 5$
$\Rightarrow y = - \dfrac{1}{4}{x^2} + 3x - 4$
Therefore, clearly, this is in the standard form $y = A{x^2} + Bx + C$
Hence, the standard form of the equation of the parabola that has the indicated vertex and whose graph passes through the given point: Vertex: $\left( {6,5} \right)$; point: $\left( {0, - 4} \right)$can be written as $y = - \dfrac{1}{4}{x^2} + 3x - 4$
Thus, 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). If we were given the standard equation, then 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.