How do you write the quadratic function #y=x^2+14x+11# in vertex form?

Question

HIX Tutor · Accepted Answer

undefined To write the quadratic function in vertex form, follow these steps:
1. Complete the square to find the vertex form.
2. Rewrite the equation y=x^2+14x+11.
3. Write the equation as y=(x^2+14x)+11.
4. To complete the square, add and subtract (14/2)^2 = 49 inside the parentheses to keep the equation balanced.
5. Rewrite the equation as y=(x^2+14x+49-49)+11.
6. Simplify to y=(x+7)^2-49+11.
7. Combine like terms to get y=(x+7)^2-38.
Therefore, the quadratic function y=x^2+14x+11 written in vertex form is y=(x+7)^2-38.

Nathan Axel · Answer

The quadratic function in vertex form is #y=(x+7)^2 -38# # y= x^2+14x+11 or y= (x+7)^2 -49+11 or y=(x+7)^2 -38# Comparing with standard vertex form #y=a(x-h)^2+k #, where #(h,k)# is vertex. So vertex is at #(-7 , -38)# The quadratic function in vertex form is #y=(x+7)^2 -38# graph{x^2+14x+11 [-160, 160, -80, 80]}[Ans]

Nathan Axel · Answer

undefined To write the quadratic function $y = x^2 + 14x + 11$ in vertex form, follow these steps:

1. Complete the square for the quadratic term:
   $y = (x^2 + 14x) + 11$
   $y = (x^2 + 14x + 49) - 49 + 11$
   $y = (x + 7)^2 - 38$

2. Rewrite the equation in vertex form:
   $y = (x + h)^2 + k$
   where $h = -7$ and $k = -38$

Therefore, the quadratic function $y = x^2 + 14x + 11$ in vertex form is $y = (x + 7)^2 - 38$.