For Question 1: a) Complete the table b) Plot the points and sketch the graph c) Label on the graph any x -intercepts, y -intercepts and vertices. You will earn: - 2 pts for correctly completing the table ( 1 pt for partially correct table) - 2 pts for plotting your points correctly and sketching a graph through them - 2 pts for labeling the key points on the graph (intercepts and vertices) 1. Given f(x)=2 x^2+8 x+4 a) Complete the table x & f(x) -4 & -3 & -2 & -1 & 0 & b) Plot the points and sketch the graph

To complete the table and sketch the graph for the function \( f(x) = 2x^2 + 8x + 4 \), we need to evaluate the function at each given \( x \)-value.<br /><br />### a) Complete the Table<br /><br />1. **For \( x = -4 \):**<br /> \[<br /> f(-4) = 2(-4)^2 + 8(-4) + 4 = 2(16) - 32 + 4 = 32 - 32 + 4 = 4<br /> \]<br /><br />2. **For \( x = -3 \):**<br /> \[<br /> f(-3) = 2(-3)^2 + 8(-3) + 4 = 2(9) - 24 + 4 = 18 - 24 + 4 = -2<br /> \]<br /><br />3. **For \( x = -2 \):**<br /> \[<br /> f(-2) = 2(-2)^2 + 8(-2) + 4 = 2(4) - 16 + 4 = 8 - 16 + 4 = -4<br /> \]<br /><br />4. **For \( x = -1 \):**<br /> \[<br /> f(-1) = 2(-1)^2 + 8(-1) + 4 = 2(1) - 8 + 4 = 2 - 8 + 4 = -2<br /> \]<br /><br />5. **For \( x = 0 \):**<br /> \[<br /> f(0) = 2(0)^2 + 8(0) + 4 = 0 + 0 + 4 = 4<br /> \]<br /><br />The completed table is:<br /><br />\[<br />\begin{array}{|c|c|}<br />\hline<br />x & f(x) \\<br />\hline<br />-4 & 4 \\<br />\hline<br />-3 & -2 \\<br />\hline<br />-2 & -4 \\<br />\hline<br />-1 & -2 \\<br />\hline<br />0 & 4 \\<br />\hline<br />\end{array}<br />\]<br /><br />### b) Plot the Points and Sketch the Graph<br /><br />Plot the points \((-4, 4)\), \((-3, -2)\), \((-2, -4)\), \((-1, -2)\), and \((0, 4)\) on a coordinate plane. Connect these points with a smooth curve to form a parabola.<br /><br />### c) Label Key Points on the Graph<br /><br />- **Vertex:** The vertex of a parabola in the form \( ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). For this function:<br /> \[<br /> x = -\frac{8}{2 \times 2} = -2<br /> \]<br /> Substituting \( x = -2 \) back into the function gives \( f(-2) = -4 \). So, the vertex is \((-2, -4)\).<br /><br />- **\( x \)-Intercepts:** Set \( f(x) = 0 \) and solve for \( x \):<br /> \[<br /> 2x^2 + 8x + 4 = 0<br /> \]<br /> Divide the entire equation by 2:<br /> \[<br /> x^2 + 4x + 2 = 0<br /> \]<br /> Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):<br /> \[<br /> x = \frac{-4 \pm \sqrt{16 - 8}}{2} = \frac{-4 \pm \sqrt{8}}{2} = \frac{-4 \pm 2\sqrt{2}}{2} = -2 \pm \sqrt{2}<br /> \]<br /> The \( x \)-intercepts are \((-2 + \sqrt{2}, 0)\) and \((-2 - \sqrt{2}, 0)\).<br /><br />- **\( y \)-Intercept:** This occurs when \( x = 0 \), so the \( y \)-intercept is \((0, 4)\).<br /><br />Label these points on your graph accordingly.