i need to find the vertex of this parabola equation y= 3x^2 -6x -17. in order to that i need to put it into

y=a(x-r)(x-s) by using the decomposition method. However it is impossible

3 x -17 = -51

_ x _ = -51

_ + _ = -6

And here is where im stuck _ x _ and _ + _ have to be the same numbers for example 2 x 5 = 10 and 2 + 5 = 7

thanks

## Answers (2)

3x^2 -6x -17 cannot be factored nicely.

3x² − 6x − 17

= 3 (x² − 2x − 17/3)

= 3 (x² − 2x + 1 − 17/3 − 1)

= 3 [ (x − 1)² − 20/3 ]

= 3(x − 1)² − 20

vertex is (1, −20).

No. You can find the vertex by computing -b/2a.

x=-b/2a=-(-6)/((2)(3)) = 1

y=3(1)Â²-6(1)-17=3-6-17=-20

So the vertex is at (1,-20)

The other way to do this is to turn it into vertex form:

y=a(x-h)Â²+k and (h,k) will be your vertex.

To do this we must complete the square:

y=3xÂ²-6x-17 First we factor out the 3 to get

y=3(xÂ²-2x)-17 Now we complete the square by adding and subtracting (b/2)Â²

y=3(xÂ²-2x+1-1)-17 We had to add and subtract 1 because this is a

expression not an equation.

Now we distribute out the -1

y=3(xÂ²-2x+1)-3-17 Now we change (xÂ²-2x+1)=(x-1)Â² and combine -3-17

y=3(x-1)Â²-20 Now we have the vertex (1,-20)