![y=3x^2-4x+1](https://img.qammunity.org/2023/formulas/mathematics/college/abt6xfhzj94h0mksc4usaomfxvdbcvt47p.png)
To find the vertex of a parabola (quadratic equiation) you use the next:
1. Find the axis of simetry (value of x in the vertex) with the next formula:
![\begin{gathered} y=ax^2+bx+c \\ \\ x=(-b)/(2a) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/nmtclkriuwu10o5povyj6862hchgsjvca8.png)
For the given equation:
b=-4
a=3
![x=(-(-4))/(2(3))=(4)/(6)=(2)/(3)](https://img.qammunity.org/2023/formulas/mathematics/college/u8iiokg2ggjanmfco81digw6g7pgf6cfr5.png)
Axis of simetry x=2/3
2. Find the value of y in the vertex. Use the axis of simetry:
![\begin{gathered} y=3((2)/(3))^2-4((2)/(3))+1 \\ \\ y=3((4)/(9))-(8)/(3)+1 \\ \\ y=(12)/(9)-(8)/(3)+1 \\ \\ y=(4)/(3)-(8)/(3)+(3)/(3) \\ \\ y=-(1)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/srp5mxxd5w590hq9kcb6o8ak1eqm2trg0t.png)
3. The vertex is (2/3 , -1/3)
4. Write in vertex form.
General vertex form of a quadratic equation:
![y=a(x-h)^2+k](https://img.qammunity.org/2023/formulas/mathematics/college/97p0xsjs0cwme4ddvwkim2cbbqprhnlhsv.png)
The vertex is (h,k)
For the given equation:
a=3
h=2/3
k= -1/3
Equation in vertex form:
![y=3(x-(2)/(3))^2-(1)/(3)](https://img.qammunity.org/2023/formulas/mathematics/college/yqdvhxf5tcwm83olppcapl0z7b5dpxkv1k.png)