# Find the vertex and zeroes of x^2-12y+5=0

1
by Maeveewaveeeeeee
The variables are different..................................

• Brainly User
2015-12-31T12:37:03+08:00
Y = ax² + bx + c    ⇒    y = f(x)
f(x) = ax² + bx + c

Given:  x² - 12y + 5 = 0
Convert to y = ax² + bx + c
x² - 12y + 5 = 0
x² + 5 = 12y
12y = x² + 5
12y/12 = x²/12 + 5/12
y =

A.)  Set y to = 0

Solve for roots (zeroes) using the method extracting the square roots.
Use this method when b = 0 in equation ax² + bx + c = 0.

x² + 5 = 0

x² = -5

x₁ =

x₂ =

THE ZEROES (ROOTS) are   and .

It means that the equation has no real roots, and the graph (parabola) that opens upward is above the x-axis.

B.)  Find the vertex of the parabola.
Since the equation has a positive leading leading term (), the parabola opens upward (u-shaped), and the vertex is the minimum.

Vertex = (h, k)

h =

h =

h = 0

k = f(h)
Plug -in  the value of h (0) to x in equation

k =

k = 0 + ⁵/₁₂

k = ⁵/₁₂

Vertex = (h, k)
Vertex = (0, ⁵/₁₂)

FINAL ANSWER:  The vertex is (0, ⁵/₁₂) and the zeroes (roots) are  and .

Please click image to see the graph of the given equation.