Question

I neex help converting this from standard form into vertex f (x)=2x^2+36x+170

Answer 1

Y=2x^2+36x+170  setup equation this way to find the parabola.

Binomial should equal zero. Disregard the constant value of 170 and replace with c. So, 0=2x^2+36x+c is your equation.
Switch your x from right to left side showing 2x^2+36x=0, divide by two and simplify.  Setup like this: 2x^2/2 + 36x/2 = 0/2.  Becomes, x2 +36x/2=0/2.
Reduce common factors to equation reads x2+18x=0/2. Divide zero by two to get 0 so, it looks like this x^2+18x=0.
Create a trinomial square on the left side of equation, you must find a value equal to the square of half of b, coefficient of x which is (b/2)^2=(9)^2 add this term to each side of the equation to get x^2+18x+(9)^2=0+(9)^2. Simplify.
x^2+18x+81=81.  Now, factor the perfect trinomial square into (x+9)^2 into (x+9)^2=81.  Move the new constant to the left side equation. Becomes, 
2(x+9)^2 - 162=0.  Now, add the original constant to the new constant to get
2(x+9)^2 - 162+170=0.  You must complete the square in the expression
2x^2+36x+170 for 2(x+9)^2 + 8.  Now, reorder the right side of the equation, matching the vertex form of a parabola which is y=2(x+9)^2+8.  You must use the vertex form y=a(x-h)^2+k to figure out the values of a, h, and k.
a=2   h=-9   k=8   the vertex of (h, k) is (-9,8).