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

Question

Ethan Adkins · Accepted Answer

The vertex form of a quadratic function is given by
#y = a(x - h)^2 + k#, where #(h, k)# is the vertex of the parabola.  We can write the given equation as # color(green)(y = 1*(x - 0)^2 + (-5)# where # a = 1; h = 0; k = -5# The vertex of the Parabola is# {0 , -5}#

Abigail Allen · Answer

undefined To write $y = x^2 - 5$ in vertex form, which is $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola, follow these steps:

Given $y = x^2 - 5$, notice that the equation is already in a form close to the vertex form, but without the complete square. For this particular equation, $a = 1$, and the equation lacks a linear $x$ term, which simplifies the process since there's no need to complete the square.

The given equation directly translates to a vertex form of $y = 1(x - 0)^2 - 5$, because:

1. The coefficient $a = 1$.
2. There is no $x$ term, which means the $h$ value that would normally shift the vertex left or right is $0$.
3. The constant term, $-5$, represents the $k$ value, shifting the vertex down.

Thus, the vertex form of the equation $y = x^2 - 5$ is $y = (x - 0)^2 - 5$, or more simply, $y = x^2 - 5$, with the vertex at $(0, -5)$.