91Ó°ÊÓ

The vertex formula is given by \[x = \frac{-b}{2a}\], where a and b are the coefficients in the quadratic function \(g(x) = ax^2 + bx + c\). In this case, \(a = 2\) and \(b = -4\). Plugging these values into the formula to find the x-coordinate of the vertex, we get \[x = \frac{-(-4)}{2(2)} = \frac{4}{4} = 1\]. To find the y-coordinate of the vertex, plug the x-coordinate back into the function: \[g(1) = 2(1)^2 - 4(1) + 4 = 2 - 4 + 4 = 2\]. Therefore, the vertex of the parabola is V(1, 2).

To find the x-intercepts, we set the function equal to zero and solve for x: \[0 = 2x^2 - 4x + 4\]. However, it's difficult to factor this equation or solve for x directly. So, let's use the quadratic formula to find the x-intercepts: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. With \(a = 2\), \(b = -4\), and \(c = 4\), we get: \[x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(4)}}{2(2)} = \frac{4 \pm \sqrt{0}}{4}\]. Since the discriminant is 0 (\(b^2 - 4ac = 0\)), there is only one x-intercept: \[x = \frac{4}{4} = 1\]. This means the x-intercept coincides with the vertex, and the point is (1,0). To find the y-intercept, we set \(x = 0\) and solve for \(g(x)\): \[g(0) = 2(0)^2 - 4(0) + 4 = 4\]. This gives us the y-intercept at the point (0, 4).

Now we have the vertex V(1, 2), the x-intercept (1, 0), and the y-intercept (0, 4). Plot these points on a graph, and then sketch the parabola using these points, remembering that the graph opens upwards (since the coefficient \(a = 2 > 0\)). This will give you the graph of the function \(g(x) = 2x^2 - 4x + 4\).