Question

How do I convert f(x) = x2 + 6x – 16 into the general, vertex form of the equation?

Answer 1

The vertex form of a parabola is written as \[y=a(x-h)^2+k\] All you need to do is find a, and the vertex. Looking at the original equation, you know that a=1. To find the vertex (h,k) you can use the formula x=-b/2a. Then, you'll get that x=-3. If you plug that back in, you'll get that y=-25. So the vertex is (-3, -25), which means that h=-3 and k=-25.
If you plug these values into the vertex form equation you'll get\[y=1(x+3)^2-25\]

Answer 2

Wow, Thank you TrainWreck! Appreciate it man.

Answer 3

@trainwrecking , would you mind helping me with another problem?
If so, how would you find the solutions of g(x) = x2 +6x + 1 ?
Thank you in advance!

Answer 4

Sure, no problem!
It seems that this equation is un-factorable, so you have to use the quadratic formula\[\frac{ -b+ \sqrt{b^2-4ac} }{ 2a}\]
\[\frac{ -b- \sqrt{b^2-4ac} }{ 2a}\]
Just substitute the values to those two expressions and you should get the answer!

\[\frac{ -6 +\sqrt{36-4}}{ 2 }\]
\[\frac{ -6 -\sqrt{36-4}}{ 2 }\]

And so your solutions are
\[x= -3+\frac{ \sqrt{34} }{ 2 } and -3-\frac{ \sqrt{34} }{ 2 }\]

Answer 5

Wow Thank you again :D.

Answer 6

Wait, I'm sorry, messed up the last part!

Answer 7

Oh alright

Answer 8

The solutions would actually be \[x= -3 + 2\sqrt{2} and -3 - 2\sqrt{2}\]

Answer 9

Oh ok, Thanks for showing work

Answer 10

No problem :)