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

Jun 2, 2017

$\left(\frac{23}{12} , \frac{767}{24}\right)$

#### 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:

$f \left(x\right) = a {x}^{2} + b x + c$

where $a , b ,$ and $c$ are constants and $\left(\frac{- b}{2 a} , f \left(\frac{- b}{2 a}\right)\right)$ is the vertex.

$y = 3 {x}^{2} + x + 6 + 3 \left({x}^{2} - 8 x + 16\right)$

$y = 3 {x}^{2} + x + 6 + 3 {x}^{2} - 24 x + 48$

$y = 6 {x}^{2} - 23 x + 54$

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

$\frac{- b}{2 a} = \frac{23}{12}$

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

$y = 6 {\left(\frac{23}{12}\right)}^{2} - 23 \left(\frac{23}{12}\right) + 54$

$y = \frac{529}{24} - \frac{529}{12} + 54$

$y = - \frac{529}{24} + \frac{54 \cdot 24}{24}$

$y = \frac{1296 - 529}{24} = \frac{767}{24}$

So the vertex is $\left(\frac{23}{12} , \frac{767}{24}\right)$