What is the vertex form of y=x^2+4x-2?

2 Answers
Jul 27, 2016

(x + 2)^2 - 6

Explanation:

First, find the coordinates of the vertex.
x-coordinate of vertex
x = -b/(2a) = -4/2 = -2
y-coordinate of vertex
y(-2) = 4 - 8 - 2 = -6
Vertex (-2, -6)
Vertex form of y:
y = (x + 2)^2 - 6

Jul 27, 2016

y=(x+2)^2-6

Explanation:

We begin with y=x^2+4x-2. In order to find the vetex form of this equation we need to factor it. If you try it, y=x^2+4x-2 is not dactorable, so now we can either complete the square or use the quadratic formula. I'm going to use the quadratic formula because it is fool-proof, but learning how to complete the square is valuable too.

The quadratic formula is x=(-b+-sqrt(b^2-4*a*c))/(2*a), where a, b, c come from ax^2 +bx+c. In our case, a=1, b =4, and c=-2.

That gives us x=(-4+-sqrt(4^2-4*1*-2))/(2*1), or (-4+-sqrt(16-(-8)))/2, which simplifies further to (-4+-sqrt(24))/2.

From here we expand sqrt(24) to 2sqrt(6), which makes the equation (-4+-2sqrt(6))/2, or -2+-sqrt(6).

So we went from x=(-4+-sqrt(4^2-4*1*-2))/(2*1) to x=-2+-sqrt(6). Now we add 2 on both sides, leaving us with +-sqrt6=x+2. From here, we need to get rid of the square root, so we'll square both sides, which will give us 6=(x+2)^2. Subtarct 6, and have 0=(x+2)^2-6. Since we're looking for the eqaution when y=0 (the x-axis), we can use 0 and y interchanagbly.

Thus, 0=(x+2)^2-6 is the same thing as y=(x+2)^2-6. Nice work, we ow have the equation in Vertex form!