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Question: How do you write \( f\left( x \right) = 1 - 6x - {x^2}\) into vertex form?...

How do you write f(x)=1−6x−x2 f\left( x \right) = 1 - 6x - {x^2} into vertex form?

Explanation

Solution

To determine the solution to the above equation using completing the square method, pull out common the coefficient of x2 {x^2} from terms containing x x as to obtain the form x2+bx {x^2} + bx and add (b2)2 {\left( {\dfrac{b}{2}} \right)^2}
(b2)2{\left( {\dfrac{b}{2}} \right)^2} to make x2+bx+(b2)2 {x^2} + bx + {\left( {\dfrac{b}{2}} \right)^2} which is a perfect square of (x+b2)2 {\left( {x + \dfrac{b}{2}} \right)^2} .Now compare the obtained equation with the vertex form f(x)=a(x−h)2+k f\left( x \right) = a{\left( {x - h} \right)^2} + k to get the required result.

Complete step by step solution:
We are given a quadratic equation f(x)=1−6x−x2 f\left( x \right) = 1 - 6x - {x^2} .
Vertex form of a quadratic equation is given by f(x)=a(x−h)2+k f\left( x \right) = a{\left( {x - h} \right)^2} + k where a a tells whether the graph of the quadratic equation is open up or open down and (h,k) \left( {h,k} \right) is the vertex of the parabola.
In order to convert the above equation into the vertex form,
First pull out common the coefficient of x2 {x^2} from the x2 {x^2} and x x terms as to obtain the form x2+bx {x^2} + bx in the equation only if the coefficient is not equal to 1.
Since in our case the coefficient is equal to −1 - 1 so pulling out common −1 - 1 from the terms containing x x
f(x)=−(x2+6x)+1f\left( x \right) = - \left( {{x^2} + 6x} \right) + 1
To make the perfect square from the above we have to add(b2)2 {\left( {\dfrac{b}{2}} \right)^2} to make x2+bx+(b2)2 {x^2} + bx + {\left( {\dfrac{b}{2}} \right)^2} which is a perfect square of (x+b2)2 {\left( {x + \dfrac{b}{2}} \right)^2} where b is the coefficient of x .
(b2)2=(62)2=(3)2=9{\left( {\dfrac{b}{2}} \right)^2} = {\left( {\dfrac{6}{2}} \right)^2} = {\left( 3 \right)^2} = 9
Therefore we have to add and subtract 9 9 in the equation to make it a square
f(x)=−(x2+6x+9)+1+9f\left( x \right) = - \left( {{x^2} + 6x + 9} \right) + 1 + 9
x2+6x+9{x^2} + 6x + 9 can be written as (x+3)2 {\left( {x + 3} \right)^2}
f(x)=−(x+3)2+10f\left( x \right) = - {\left( {x + 3} \right)^2} + 10
Now Comparing the above equation with the vertex form of quadratic equation i.e. f(x)=a(x−h)2+k f\left( x \right) = a{\left( {x - h} \right)^2} + k , we get the value of variables ass
a=−1 (h,k)=(−3,10)  a = - 1 \\\ \left( {h,k} \right) = \left( { - 3,10} \right) \\\
Therefore, the vertex form of the given quadratic equation f(x)=1−6x−x2 f\left( x \right) = 1 - 6x - {x^2} is equal to f(x)=−(x+3)2+10 f\left( x \right) = - {\left( {x + 3} \right)^2} + 10

Additional Information:
Quadratic Equation: A quadratic equation is a equation which can be represented in the form of ax2+bx+c a{x^2} + bx + c where x x is the unknown variable and a,b,c are the numbers known where a≠0 a \ne 0 .If a=0 a = 0 then the equation will become linear equation and will no more quadratic .
The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: D=b2−4ac D = {b^2} - 4ac
Using Discriminant, we can find out the nature of the roots
If D is equal to zero, then both of the roots will be the same and real.
If D is a positive number then, both of the roots are real solutions.
If D is a negative number, then the root are the pair of complex solutions

Note: 1. One must be careful while calculating the answer as calculation error may come.
2.Vertex form conversion is always necessary when you want to predict the type of graph of the equation without actual plotting.
3. The graph of every quadratic equation is always a parabola. Since in our case we got a a negative it means the graph is open down.