5. y=(1)/(3)(x-4)^2 6. y=-2(x+3)^2+7 Axis of Symmetry: Vertex: Domain: Range: Transformations: Axis of Symmetry: Vertex: Domain: Range: Transformations:

Let's address each question step by step.Question 5: \(y=\frac{1}{3}(x-4)^{2}\)Step 1: Complete the table of values for the given parabola equation.To find the y-values, we will substitute various x-values into the equation.Let's choose x-values around the vertex (4, 0) for simplicity:- x = 2: \(y=\frac{1}{3}(2-4)^{2} = \frac{1}{3}(-2)^{2} = \frac{1}{3}(4) = \frac{4}{3}\)- x = 3: \(y=\frac{1}{3}(3-4)^{2} = \frac{1}{3}(-1)^{2} = \frac{1}{3}(1) = \frac{1}{3}\)- x = 4: \(y=\frac{1}{3}(4-4)^{2} = \frac{1}{3}(0)^{2} = 0\)- x = 5: \(y=\frac{1}{3}(5-4)^{2} = \frac{1}{3}(1)^{2} = \frac{1}{3}(1) = \frac{1}{3}\)- x = 6: \(y=\frac{1}{3}(6-4)^{2} = \frac{1}{3}(2)^{2} = \frac{1}{3}(4) = \frac{4}{3}\)The completed table:\begin{array}{|l|l|}\hlinex\ &\ y\ \\\hline\ 2\ &\ \frac{4}{3}\ \\\hline\ 3\ &\ \frac{1}{3}\ \\\hline\ 4\ &\ 0\ \\\hline\ 5\ &\ \frac{1}{3}\ \\\hline\ 6\ &\ \frac{4}{3}\ \\\hline\end{array}Step 2: Identify the axis of symmetry.The axis of symmetry is the vertical line that passes through the vertex of the parabola. For the equation \(y=\frac{1}{3}(x-4)^{2}\), the axis of symmetry is x = 4.Step 3: Identify the vertex.The vertex of the parabola is the point (h, k) where the equation is in the form \(y=a(x-h)^{2}+k\). For the given equation, the vertex is (4, 0).Step 4: Identify the domain.The domain of any parabola is all real numbers, so the domain is (-∞, ∞).Step 5: Identify the range.Since the parabola opens upwards (a positive coefficient of \(\frac{1}{3}\)), the range is from the y-coordinate of the vertex to infinity. Therefore, the range is [0, ∞).Step 6: Identify the transformations.The parabola is a vertical stretch by a factor of \(\frac{1}{3}\) and a horizontal shift to the right by 4 units from the standard parabola \(y=x^{2}\).Axis of Symmetry: x = 4Vertex: (4, 0)Domain: (-∞, ∞)Range: [0, ∞)Transformations: Vertical stretch by \(\frac{1}{3}\), horizontal shift right by 4 units.Question 6: \(y=-2(x+3)^{2}+7\)Step 1: Complete the table of values for the given parabola equation.Let's choose x-values around the vertex (-3, 7) for simplicity:- x = -5: \(y=-2(-5+3)^{2}+7 = -2(-2)^{2}+7 = -2(4)+7 = -8+7 = -1\)- x = -4: \(y=-2(-4+3)^{2}+7 = -2(-1)^{2}+7 = -2(1)+7 = -2+7 = 5\)- x = -3: \(y=-2(-3+3)^{2}+7 = -2(0)^{2}+7 = 0+7 = 7\)- x = -2: \(y=-2(-2+3)^{2}+7 = -2(1)^{2}+7 = -2(1)+7 = -2+7 = 5\)- x = -1: \( y=-2(-1