Question

Question #35cd4

Answer 1

See the explanation

Explanation:

You have not fully explained the situation from which you wish to obtain `-b/(2a)`. However, I think I know what you are after.

It is part of the process called 'completing the square' so I will 'walk you through' the process.

Consider the standardised equation format of `y=ax^2+bx+c`
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`color(blue)("The process of completing the square")`

Factor out a from part of it:

write as `y=a(x^2+b/a x)+c`

`color(brown)("................................................................")`
`color(brown)("explaining why we divide by "a" in "b/ax")`
To return this into the original equation form you would multiply EVERYTHING inside the bracket by `a`

`axxx^2=ax^2` which matches the original

`axxb/ax" "` which is the same as `" "a/axxbx`
This in turn is the same as `1xxbx=bx` which matches the original equation.
`color(brown)("................................................................")`

Take the power of 2 from `x^2` and put it outside the brackets
However, the changes in what follows introduces an error. So we compensate for this error by introducing the correction facto of `k`

`y=a(x+b/ax)^2+k+c`

Remove the `x` from `b/ax`

`y=a(x+b/a)^2+k+c`

All you need to do now if determine the value of `k`
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`color(blue)("Now comes your question part")`

The vertex (tip of the rounded bottom or top of the curve) as an `x` component and a `y` component

`x_("vertex")=color(red)(-1/2)xxb/a` this is the same as `-b/(2a)`

You just have to know that the `color(red)(-1/2xx)` is what is always needed to make the system work. You always have to do it

Substitute values and you have almost got the whole thing.
There is a bit more manipulation to determine the value of `k` but you have not asked about that.
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`color(blue)("An example")`

Suppose that we were given the values `y=3x^2+2x-5`

`a=+3"; "b=+2`

so `-b/(2a)=-((+2)/(2xx3)) = -1/3`