Question

What is the vertex of `y= 3x^2+x+6+3(x-4)^2`?

Answer 1

`(23/12, 767/24)`

Explanation:

Hmm... this parabola isn't in standard form or vertex form. Our best bet to solve this problem is to expand everything and write the equation in the standard form:

`f(x) = ax^2+bx+c`

where `a,b,` and `c` are constants and `((-b)/(2a), f((-b)/(2a)))` is the vertex.

`y = 3x^2+x+6+3(x^2-8x+16)`

`y = 3x^2+x+6+3x^2-24x+48`

`y = 6x^2-23x+54`

Now we have the parabola in standard form, where `a=6` and `b=-23`, so the `x` coordinate of the vertex is:

`(-b)/(2a) = 23/12`

Finally, we need to plug this `x` value back into the equation to find the `y` value of the vertex.

`y = 6(23/12)^2-23(23/12)+54`

`y = 529/24 - 529/12 + 54`

`y = -529/24 + (54*24)/24`

`y = (1296-529)/24 = 767/24`

So the vertex is `(23/12, 767/24)`

Final Answer