# What is the vertex of y=1/3(7x-2)^2-7? Thank you so much, in advance.?

Dec 24, 2015

Compare with the vertex form and get the answer.

#### Explanation:

$y = \frac{1}{3} {\left(7 x - 2\right)}^{2} - 7$

The vertex form would be $y = a {\left(x - h\right)}^{2} + k$ where (h,k) is the vertex.

We can write the given equation in the vertex form and get the vertex.

$y = \frac{1}{3} {\left(7 \left(x - \frac{2}{7}\right)\right)}^{2} - 7$
$y = \frac{1}{3} \left({7}^{2}\right) {\left(x - \frac{2}{7}\right)}^{2} - 7$
$y = \frac{49}{3} {\left(x - \frac{2}{7}\right)}^{2} - 7$
Now we have got it to a form which we can recognize.
Comparing with $a {\left(x - h\right)}^{2} + k$ we can see $h = \frac{2}{7} \mathmr{and} k = - 7$

The vertex is $\left(\frac{2}{7} , - 7\right)$

Alternate Method.
The alternate method is when you put $7 x - 2 = 0$ and solve for x to find $x = \frac{2}{7}$ and get x-coordinate of the vertex. When you substitute $x = \frac{2}{7}$ in the given equation you would get $y = - 7$ which would be the y-coordinate of the vertex and still you would get the vertex $\left(\frac{2}{7} , - 7\right)$