The equation is in the form $y = ax^2 + bx + c$, where $a = 2$, $b = -16$, and $c = 33$. The x-coordinate of the vertex is given by $x = -\frac{b}{2a}$.
$x = -\frac{-16}{2(2)} = \frac{16}{4} = 4$
Now substitute $x = 4$ into the equation to find the y-coordinate of the vertex:
$y = 2(4)^2 - 16(4) + 33$
$y = 2(16) - 64 + 33$
$y = 32 - 64 + 33$
$y = 1$
So, the vertex is $(4, 1)$.
The y-intercept occurs when $x = 0$. Substitute $x = 0$ into the equation:
$y = 2(0)^2 - 16(0) + 33$
$y = 33$
So, the y-intercept is $(0, 33)$.
Since the parabola opens upwards ($a > 0$) and we have the vertex $(4, 1)$ and y-intercept $(0, 33)$, we can sketch the parabola. Another point can be found by using the symmetry of the parabola across the vertical line x = 4. If one goes 4 to the left of x=4 and lands on x=0, y is 33, so four to the right of x=4 is x = 8, so one can graph the point (8,33).
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