How do I convert the equation #f(x)=x^2+2/5x−1# to vertex form?

1 Answer
Jul 29, 2015

Answer:

#color(red)( f(x) = (x+1/5)^2-26/25)#

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

#f(x) = x^2+2/5x-1#

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

Step 1. Move the constant to the other side.

#f(x)+1 = x^2+2/5x#

Step 2. Square the coefficient of #x# and divide by 4.

#(2/5)^2/4 = (4/25)/4 = 1/25#

Step 3. Add this value to each side

#f(x)+1+1/25 = x^2+2/5x+1/25#

Step 4. Combine terms.

#f(x)+26/25 = x^2+2/5x+1/25#

Step 5. Express the right hand side as a square.

#f(x)+26/25 = (x+1/5)^2#

Step 5. Isolate #f(x)#.

#f(x) = (x+1/5)^2-26/25#

The equation is now in vertex form.

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

#h = -1/5# and #k = -26/25#, so the vertex is at (#-1/5,-26/25#)

Graph