To find the vertex form of parabola [tex]y=a(x-x_0)^2+y_0[/tex] given in almost standard form [tex]y=ax^2+bx+c,[/tex] were written following steps:
1. Write the function in standard form:
[tex]y=3x^2+18x.[/tex]
2. Factor a out of the first two terms:
[tex]y=3(x^2+6x).[/tex]
3. Form a perfect square trinomial:
[tex]y=3(x^2+6x+9-9)=3(x^2+6x+9)-3\cdot 9.[/tex]
4. Write the trinomial as a binomial squared:
[tex]y=3(x+3)^2-27.[/tex]
The vertex is (-3,-27).
Answer: missing value is 3
The function in vertex form is y = 3( x+3)² -9
A function is a mathematical statement formed for relating a dependent and in independent variable.
It is given that
f(x) = 18x +3x²
To write it into vertex form
The standard vertex form is given by
y =a(x- h)² +k
y = 18x +3x²
y = 3(x² +6x)
y = 3 (x² + 2.3 x + 9) -9
y = 3( x+3)² -9
Therefore in vertex form the function is y = 3( x+3)² -9
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