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Question: How do you write the vertex form the equation of the parabola \(y=-{{x}^{2}}+12x-4\) ?...

How do you write the vertex form the equation of the parabola y=−x2+12x−4y=-{{x}^{2}}+12x-4 ?

Explanation

Solution

The vertex equation of parabola y=ax2+bx+cy=a{{x}^{2}}+bx+c is y=a(x−d)2+ey=a{{\left( x-d \right)}^{2}}+e where (d,e) is the
Coordinate of vertex of the parabola to convert the equation into vertex form we have to convert the quadratic equation of parabola into complete square form.

Complete step by step answer:
The given equation of the parabola is y=−x2+12x−4y=-{{x}^{2}}+12x-4
We have to convert the equation to vertex form of equation that is y=a(x−d)2+ey=a{{\left( x-d \right)}^{2}}+e
For that we have to convert the equation to complete square form
y=−x2+12x−4y=-{{x}^{2}}+12x-4
In RHS take a common except for the constant term , a is the coefficient of x2{{x}^{2}} in the equation
So we have take -1 common in the equation except for the constant term
So y=−1(x2−12x)−4y=-1\left( {{x}^{2}}-12x \right)-4
Now observe the coefficient of x and divide it by 2 and add the square of result inside the the bracket and add the corresponding number in LHS
So we have to add 36 inside the bracket in RHS and add -36 in LHS
y−36=−1(x2−12x+36)−4y-36=-1\left( {{x}^{2}}-12x+36 \right)-4
Now we can see that the term inside the bracket is a square term , so we can write
y−36=−1(x−6)2−4y-36=-1{{\left( x-6 \right)}^{2}}-4
We can now add 36 in both LHS and RHS
y=−1(x−6)2+32y=-1{{\left( x-6 \right)}^{2}}+32
Now the above equation is in vertex form of the equation. The coordinate of vertex of the parabola is (6, 32)

Note:
There are shorter ways to write the vertex form the equation in just one step. If we have equation of parabola in the form of y=ax2+bx+cy=a{{x}^{2}}+bx+c then the vertex for equation is y=a(x−d)2+ey=a{{\left( x-d \right)}^{2}}+e where the value of d is −b2a-\dfrac{b}{2a} and the value of e is 4ac−b24a\dfrac{4ac-{{b}^{2}}}{4a} so we can write
y=a(x−(−b2a))2+4ac−b24ay=a{{\left( x-\left( -\dfrac{b}{2a} \right) \right)}^{2}}+\dfrac{4ac-{{b}^{2}}}{4a} so the coordinate of vertex is (−b2a,4ac−b24a)\left( -\dfrac{b}{2a},\dfrac{4ac-{{b}^{2}}}{4a} \right)