Answers

  • 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 =  \frac{ x^{2} }{12}  +  \frac{5}{12}

A.)  Set y to = 0
       \frac{ x^{2} }{12} + \frac{5}{12} =0

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

      12( \frac{ x^{2} }{12} + \frac{5}{12} =0)

      x² + 5 = 0

      x² = -5

       \sqrt{ x^{2} } = \frac{+}{-}  \sqrt{-5}

      x₁ = i \sqrt{5}

      x₂ = -i \sqrt{5}

THE ZEROES (ROOTS) are i \sqrt{5}  and -i \sqrt{5} .

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 ( \frac{ x^{2} }{12} ), the parabola opens upward (u-shaped), and the vertex is the minimum.

Vertex = (h, k)

h =  \frac{-b}{2a}  

h =  \frac{0}{2( \frac{1}{12}) }

h = 0

k = f(h)
Plug -in  the value of h (0) to x in equation   \frac{x ^{2} }{12} + \frac{5}{12}

k =  \frac{0 ^{2} }{12} + \frac{5}{12}

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 i \sqrt{5} and -i \sqrt{5} .

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