How do you write the quadratic function #y=2x^2+4x-5# in vertex form?

2 Answers
smendyka
Jul 24, 2017

See a solution process below:

Explanation:

To convert a quadratic from #y = ax^2 + bx + c# form to vertex form, #y = a(x - color(red)(h))^2+ color(blue)(k)#, you use the process of completing the square.

First, we must isolate the #x# terms:

#y + color(red)(5) = 2x^2 + 4x - 5 + color(red)(5)#

#y + 5 = 2x^2 + 4x#

We need a leading coefficient of #1# for completing the square, so factor out the current leading coefficient of 2.

#y + 5 = 2(x^2 + 2x)#

Next, we need to add the correct number to both sides of the equation to create a perfect square. However, because the number will be placed inside the parenthesis on the right side we must factor it by #2# on the left side of the equation. This is the coefficient we factored out in the previous step.

#y + 5 + (2 * ?) = 2(x^2 + 2x + ?)#

#y + 5 + (2 * 1) = 2(x^2 + 2x + 1)#

#y + 5 + 2 = 2(x^2 + 2x + 1)#

#y + 7 = 2(x^2 + 2x + 1)#

Then, we need to create the square on the right hand side of the equation:

#y + 7 = 2(x + 1)^2#

Now, isolate the #y# term:

#y + 7 - color(red)(7) = 2(x + 1)^2 - color(red)(7)#

#y + 0 = 2(x + 1)^2 - 7#

#y = 2(x + color(red)(1))^2 - color(blue)(7)#

Or, in precise form:

#y = 2(x - color(red)((-1)))^2 + color(blue)((-7))#

The vertex is: #(-1, -7)#

Nghi N
Jul 25, 2017

#y = 2(x + 1)^2 - 7#

Explanation:

#y = 2x^2 + 4x - 5#
x-coordinate of vertex:
#x = -b/(2a) = -4/4 = - 1#
y-coordinate of vertex:
#y(-1) = 2(-1)^2 +4(-1) - 5 = 2 - 4 - 5 = - 7#
Vertex (-1, -7)
Vertex form of y:
#y = 2(x + 1)^2 - 7#

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