Originally Posted by
masters What you did was solve for the x-intercepts. This is where y=0. This is the smallest value for y only when the parabola is tangent to the x-axis, as in this case. The vertex is on the x-axis at (4, 0)
If the vertex had been below or above the x-axis, setting the equation equal to zero and solving will only find you the zeros (x-intercepts), not the minimum point.
To find the x-coordinate of the minimum/maximum point use
$\displaystyle x=\frac{-b}{2a}$ from $\displaystyle f(x)=ax^2+bx+c$
Then, substitute that into your original equation to find the y-coordinate. Thus, the minimum/maximum point would be
$\displaystyle \left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$