The given form is the general form of the quadratic:
`f(x) = ax^2+bx+c`
The standard form (also called the vertex form) is:
`f(x) = a(x-h)^2+k` where h and k are the co-ordinates of teh vertex.
To convert `2x^2+6x-20` to standard form:
1. Factor out the co-efficient of `x^2` from the x-es only, the co-efficient being the number in front.
`therefore f(x) = 2x^2+6x -20` becomes `f(x) = 2(x^2 +6/2x) -20`
`therefore f(x) = 2(x^2+3x) -20`
Now complete the square. To do this take half the middle term but not using the x from 3x (ie 3 divide by 2 = `3/2` ) and apply it inside a bracket with one of your x-es from `x^2` . Show the square outside the bracket ie `(x+3/2)^2` . Now, you have to minus what you have just added to make the equation make sense just as +1 -1 makes sense and does not change the essence. So we have `- 2(3/2)^2` to counteract the other 3/2 which appears in the first bracket. Note to include the 2 that we initially factored out as well to make the it identical :
`f(x) =[2(x +3/2)^2] -20 - 2(3/2)^2`
`therefore f(x) = 2(x+3/2)^2 - 24 1/2`
This reflects the standard form.
Ans: f(x) = 2(x+`3/2` `)^2` - 24 `1/2`