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

Question

Ethan Adkins · Accepted Answer

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

Eva Abramson · Answer

See a solution process below:

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)#

Eva Adamson · Answer

undefined To write the quadratic function $y = 2x^2 + 4x - 5$ in vertex form, follow these steps:

1. Factor out the leading coefficient from the $x^2$ and $x$ terms.
2. Complete the square for the quadratic term.
3. Rewrite the equation in vertex form by combining the completed square term with the constant term.
4. The vertex form of the quadratic function is $y = a(x - h)^2 + k$, where $(h, k)$ represents the coordinates of the vertex.

Applying these steps:

1. Factor out the leading coefficient $2$ from the $x^2$ and $x$ terms: 
$$ y = 2(x^2 + 2x) - 5 $$

2. Complete the square for the quadratic term:
$$ y = 2(x^2 + 2x + 1 - 1) - 5 $$

3. Rewrite the equation in vertex form:
$$ y = 2(x^2 + 2x + 1) - 2 - 5 $$
$$ y = 2(x + 1)^2 - 7 $$

So, the quadratic function $y = 2x^2 + 4x - 5$ in vertex form is $y = 2(x + 1)^2 - 7$.