how do you find out the vertex when graphing inequalities of quadratics in equations like this y<x^2-16
it factors out to like x+4 x-4 and I know that i can use that to find the x intercepts but the way to find the vertex i don't understand
To find the vertex of a quadratic graph represented by an inequality equation, such as y < x^2 - 16, you can follow these steps:
Step 1: Rewrite the equation in vertex form.
The vertex form of a quadratic equation is given by y = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex.
To rewrite the equation y < x^2 - 16 in vertex form, we need to complete the square. Let's start by isolating the quadratic term on the right side of the inequality:
x^2 - 16 = y
Step 2: Add and subtract a value to complete the square.
In this case, we need to add and subtract a middle term to complete the square. Since the coefficient of x^2 is 1, we take half of the coefficient of x (which is 1) and square it:
x^2 - 16 = y
x^2 - 16 + 16 = y + 16
x^2 = y + 16
Step 3: Rewrite the equation in vertex form.
Now we can rewrite the equation in vertex form by factoring the perfect square:
x^2 = y + 16
x^2 = 1(y + 16)
By comparing this equation to the vertex form equation, we can infer that a = 1, h = 0 (since x - 0 = x), and k = 16.
Therefore, the vertex of this quadratic graph is at the point (h, k) = (0, 16).