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 = [tex] \frac{ x^{2} }{12} + \frac{5}{12} [/tex]
A.) Set y to = 0
[tex] \frac{ x^{2} }{12} + \frac{5}{12} =0[/tex]
Solve for roots (zeroes) using the method extracting the square roots.
Use this method when b = 0 in equation ax² + bx + c = 0.
[tex]12( \frac{ x^{2} }{12} + \frac{5}{12} =0)[/tex]
x² + 5 = 0
x² = -5
[tex] \sqrt{ x^{2} } = \frac{+}{-} \sqrt{-5} [/tex]
x₁ = [tex]i \sqrt{5} [/tex]
x₂ = [tex]-i \sqrt{5} [/tex]
THE ZEROES (ROOTS) are [tex]i \sqrt{5} [/tex] and [tex]-i \sqrt{5} [/tex].
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 ([tex] \frac{ x^{2} }{12} [/tex]), the parabola opens upward (u-shaped), and the vertex is the minimum.
Vertex = (h, k)
h = [tex] \frac{-b}{2a} [/tex]
h = [tex] \frac{0}{2( \frac{1}{12}) } [/tex]
h = 0
k = f(h)
Plug -in the value of h (0) to x in equation [tex] \frac{x ^{2} }{12} + \frac{5}{12} [/tex]
k = [tex] \frac{0 ^{2} }{12} + \frac{5}{12} [/tex]
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 [tex]i \sqrt{5} [/tex] and [tex]-i \sqrt{5} [/tex].
Please click image to see the graph of the given equation.
