How do you write #4y= x^2 - 2x - 31# into vertex form?

1 Answer
Jul 29, 2015

#color(red)(y = 1/4(x-1)^2-8)#

Explanation:

The vertex form of a quadratic is given by #y = a(x – h)^2 + k#, where (#h, k#) is the vertex.

The "#a#" in the vertex form is the same "#a#" as in #y = ax^2 + bx + c#.

Your equation is

#4y = x^2-2x-31#

Step 1. Divide both sides by the coefficient of #y#.

#y = 1/4x^2-1/2x-31/4#

We convert to the "vertex form" by completing the square.

Step 2. Move the constant to the other side.

#y+31/4 = 1/4x^2-1/2x#

Step 3. Factor out the coefficient #a#,

#y+31/4 = 1/4(x^2-2x)#

Step 4. Square the new coefficient of #x# and divide by 4.

#(-2)^2/4 = 1#

Step 5. Add and subtract this value inside the parentheses..

#y+31/4 = 1/4(x^2-2x +1-1)#

Step 6. Express the right hand side as a square.

#y+31/4 = 1/4((x-1)^2-1)#

Step 7. Distribute.

#y+31/4 = 1/4(x-1)^2-1/4×1#

Step 8. Multiply.

#y+31/4 = 1/4(x-1)^2-1/4#

Step 9. Isolate #y#.

#y = 1/4(x-1)^2-1/4-31/4#

Step 10. Combine like terms.

#y = 1/4(x-1)^2-32/4#

#y = 1/4(x-1)^2-8#

The equation is now in vertex form.

#y = a(x – h)^2 + k#, where (#h, k#) is the vertex.

#h = 1# and #k = -8#, so the vertex is at (#1,-8#).

Graph 1