Question

Complete the square

to find the vertex
of this parabola.
2.
х
- 16 Y-4 x-12=0

Answer 1

Vertex coordinates; (h,k) = (2, -1)

Explanation:

Vertex form of a parabola is;

y = a(x - h)² + k

Now, we are given;

x² - 16y - 4x - 12 = 0

Rearranging, we have;

x² - 4x = 16y + 12

Let's complete the square on the left hand side;

First add square of half of the coefficient of x to both sides;

x² - 4x + (-½ × 4)² = 16y + 12 + (-½ × 4)²

x² - 4x + 4 = 16y + 12 + 4

LHS can be expressed as;

(x - 2)² = 16y + 16

16y = (x - 2)² - 16

Divide both sides by 16 to get;

y = (1/16)(x - 2)² - 1

Comparing this with y = a(x - h)² + k, we have;

a = 1/16

h = 2

k = -1

(h,k) = (2, -1)