Question

What is the vertex form of `y=2x^2+11x+12`?

Answer 1

Yhe vertex form is `y=2(x+11/4)^2-25/8`

Explanation:

To find the vertex form, you complete the square

`y=2x^2+11x+12`

`y=2(x^2+11/2x)+12`

`y=2(x^2+11/2x+121/16)+12-121/8`

`y=2(x+11/4)^2-25/8`

The vertex is `=(-11/4, -25/8)`

The symmetry line is `x=-11/4`

graph{(y-(2x^2+11x+12))(y-1000(x+11/4))=0 [-9.7, 2.79, -4.665, 1.58]}

Answer 2

`color(blue)(y=2(x+11/4)^2-25/8)`

Explanation:

Consider the standardised form of `y=ax^2+bx+c`

The vertex form is: `y=a(x+b/(2a))^2+k+c`

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`color(brown)("Additional note about the method")`
By rewriting the equation in this form you introduce an error. Let me explain.

Multiply out the bracket in `y=a(x+b/(2a))^2+c` and you get:

`y=a[x^2+(2xb)/(2a)+(b/(2a))^2]+c`

`color(green)(y=ax^2+bx+color(red)(a(b/(2a))^2)+c)`

the `color(red)(a(b/(2a))^2)` is not in the original equation so it is the error. Thus we need to 'get rid' of it. By introducing the correction factor of `k` and setting `color(red)(a(b/(2a))^2+k=0)` we 'force' the vertex form back into the value of the original equation.
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Given:`" "y=ax^2+bx+c" "->" "y=2x^2+11x+12`

`y=a(x+b/(2a))^2+k+c" "->" "y=2(x+11/4)^2+k+12`

But:
`a(b/(2a))^2+k=0" "->" "2(11/4)^2+k=0`

`=>k=-121/8`

So by substitution we have:
`y=a(x+b/(2a))^2+k+c" "->y=2(x+11/4)^2-121/8+12`

`color(blue)(y=2(x+11/4)^2-25/8)`
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The two equation have been plotted to show that they produce the same curve. One is thicker than the other so that they can both be seen.

Tony B