How would these graphs be sketched \( y=3 x^{2}+6 x+1 \) \( y=x^{2}-6 x+9 \) \( y=2 x^{2}+6 x+10 \)

Below is one way to sketch each quadratic graph. In general, you want to identify key features such as the vertex, the axis of symmetry, y‐intercept, and (if possible) x–intercepts. Completing the square is an effective method here. ────────────────────────── 1) For y = 3x² + 6x + 1 Step 1. Rewrite in vertex form by completing the square: • Factor 3 from the quadratic and linear terms:   y = 3(x² + 2x) + 1 • Complete the square inside the parenthesis:   x² + 2x = (x + 1)² – 1 • Substitute back:   y = 3[(x + 1)² – 1] + 1 = 3(x + 1)² – 3 + 1 = 3(x + 1)² – 2 Thus, the vertex is (–1, –2). Step 2. Determine the axis of symmetry:   x = –1 Step 3. Find the y-intercept by setting x = 0:   y = 3(0)² + 6(0) + 1 = 1   Point: (0, 1) Step 4. Find the x-intercepts by solving 3x² + 6x + 1 = 0:   Use the quadratic formula:    x = [–6 ± √(6² – 4·3·1)] / (2·3) = [–6 ± √(36 – 12)] / 6 = [–6 ± √24] / 6    Simplify: √24 = 2√6, so    x = [–6 ± 2√6] / 6 = –1 ± (√6)/3   Thus, the x-intercepts are at:    x = –1 + (√6)/3 and x = –1 – (√6)/3 Step 5. Sketch the graph:   • Plot the vertex (–1, –2).   • Draw the axis x = –1.   • Mark the intercepts: (0,1) on the y-axis and the two x–intercepts found.   • Since the coefficient 3 is positive, the parabola opens upward.   • Connect the points with a smooth, symmetric curve. ────────────────────────── 2) For y = x² – 6x + 9 Step 1. Recognize that the quadratic factors nicely:   x² – 6x + 9 = (x – 3)²   Thus, the vertex form is already:    y = (x – 3)² Step 2. Determine the key features:   • Vertex: (3, 0)   • Axis of symmetry: x = 3 Step 3. Find the y-intercept by setting x = 0:   y = (0 – 3)² = 9   Point: (0, 9) Step 4. Find the x-intercepts:   Set y = 0: (x – 3)² = 0, so x = 3. (This is a double root, meaning the graph just touches the x–axis here.) Step 5. Sketch the graph:   • Plot the vertex (3, 0).   • Draw the axis x = 3.   • Mark the y-intercept at (0, 9).   • Since the parabola opens upward (coefficient 1 > 0) and there is only one x–intercept (a “touching” point), draw the U–shaped curve accordingly. ────────────────────────── 3) For y = 2x² + 6x + 10 Step 1. Complete the square:   Factor 2 from the x-terms:    y = 2(x² + 3x) + 10   Complete the square for (x² + 3x):    x² + 3x = (x + 3/2)² – (9/4)   Now, substitute:    y = 2[(x + 3/2)² – 9/4] + 10 = 2(x + 3/2)² – (9/2) + 10    Convert 10 into halves: 10 = 20/2, so:    y = 2(x + 3/2)² + (11/2) Thus, the vertex is (–3/2, 11/2) or (–1.5, 5.5). Step 2. Determine the axis of symmetry:   x = –3/2 Step 3. Find the y-intercept (x = 0):   y = 2(0)² + 6(0) + 10 = 10   Point: (0, 10) Step 4. Check for x-intercepts by solving 2x² + 6x + 10 = 0:   Compute the discriminant:    D = 6² – 4·2·10 = 36 – 80 = –44   Since D < 0, there are no real x–intercepts (the graph does not cross the x–axis). Step 5. Sketch the graph:   • Plot the vertex (–1.5, 5.5).   • Draw the axis of symmetry x = –1.5.   • Mark the y–intercept at (0, 10).   • Since the coefficient 2 is positive, the parabola opens upward.   • Note that because there are no x-intercepts, the entire parabola stays above the x–axis.   • Draw a smooth, U–shaped curve accordingly. ────────────────────────── Summary of Sketching Steps: • Rewrite each quadratic in vertex form to determine the vertex and axis of symmetry. • Determine intercepts by setting x = 0 (for y–intercept) and solving the equation for x (x–intercepts), if possible. • Note the direction (upward or downward) from the sign of the quadratic coefficient. • Plot the vertex, intercepts, and a few additional points if necessary. • Draw a smooth parabolic curve passing through these points with symmetry about the axis. Following these steps for each equation gives you a clear sketch of the graph.