Respuesta :
Answer:
[tex]y=\frac{1}{2}(x-3)^2+\frac{1}{2}[/tex]Explanation:
The vertex form of a quadratic equation is generally given as;
[tex]y=a(x-h)^2+k[/tex]where (h, k) represents the coordinate of the vertex of the parabola.
Given a vertical intercept of 5, this means that x = 0 at y = 5. We're also given that the function has a minimum value at the point (3, 1/2), this means that h = 3 and k = 1/2.
Let's go ahead and substitute the above values into the vertex equation and solve for a;
[tex]\begin{gathered} 5=a(0-3)^2+\frac{1}{2} \\ 5=9a+\frac{1}{2} \end{gathered}[/tex]Let's subtract 1/2 from both sides of the equation;
[tex]\begin{gathered} 5-\frac{1}{2}=9a \\ \frac{10-1}{2}=9a \\ \frac{9}{2}=9a \end{gathered}[/tex]Let's now divide both sides by 9;
[tex]\begin{gathered} \frac{9a}{9}=\frac{\frac{9}{2}}{9} \\ a=\frac{9}{2}\cdot\frac{1}{9} \\ a=\frac{1}{2} \end{gathered}[/tex]Since a = 1/2, h = 3, and k = 1/2, we can go ahead and write the required formula in vertex form as;
[tex]y=\frac{1}{2}(x-3)^2+\frac{1}{2}[/tex]