Question

How do you solve \[F(x) = 3{x^2} - 30x + 82\] into vertex form?

Answer 1

Hint: To solve this question, we have to recall the vertex form of an equation. Then, we use the given terms and manipulate them to apply the identity \[{(a - b)^2} = {a^2} - 2ab + {b^2}\] in the equation, such that the remaining terms can be simplified to obtain the vertex form of the equation.

Complete step by step solution:
The vertex form of an equation is \[a{(x - h)^2} + k\] .
We have the given expression as \[F(x) = 3{x^2} - 30x + 82\] .
On analysing the expression, we notice that the identity \[{(a - b)^2} = {a^2} - 2ab + {b^2}\] can be applied here, after adjusting the expression. To achieve this form,
We need to first take out a common factor for all of the term. We can do that as follows:
  \[F(x) = 3\left( {{x^2} - 10x + \dfrac{{82}}{3}} \right)\]
Now, for identity, we need to split \[\dfrac{{82}}{3}\] in two parts such that the value of one part is \[25\] .
We can do that in this way, and simplify the expression further to obtain the vertex form of the equation.
  \[
  F(x) = 3\left( {{x^2} - 10x + \dfrac{{75 + 7}}{3}} \right) \\
   \Rightarrow F(x) = 3\left( {{x^2} - 10x + 25 + \dfrac{7}{3}} \right) \\
   \Rightarrow F(x) = 3\left( {{x^2} - 10x + 25} \right) + 7 \\
   \Rightarrow F(x) = 3{\left( {x - 5} \right)^2} + 7 \;
 \]
Hence, the vertex form of \[F(x) = 3{x^2} - 30x + 82\] is \[F(x) = 3{\left( {x - 5} \right)^2} + 7\] .
So, the correct answer is “ \[F(x) = 3{\left( {x - 5} \right)^2} + 7\] ”.

Note: The Vertex form is simply another form of a quadratic equation. The standard form of a quadratic equation is \[a{x^2} + bx + c\] , while the vertex form of a quadratic equation is \[a{(x - h)^2} + k\] . Here, \[a\] is a constant that tells us whether the parabola opens upwards or downwards, and \[(h,k)\] is the location of the vertex of the parabola. We cannot immediately read this from the standard form of a quadratic equation. Thus, vertex form is useful for solving quadratic equations, graphing quadratic functions, and more.