To sketch the graph of $y = x^2 + 2x + 3$, let’s analyze the quadratic equation step-by-step:

1. Identify the general form of the quadratic equation

The equation $y = x^2 + 2x + 3$ is in standard form:
$y = ax^2 + bx + c$ Here:

• $a = 1$ (coefficient of $x^2$)
• $b = 2$ (coefficient of $x$)
• $c = 3$ (constant term).

Since $a > 0$, the parabola opens upward.

2. Find the vertex of the parabola

The vertex $(h, k)$ of a parabola in standard form can be found using:
$x = -\frac{b}{2a}$ Substitute $a = 1$ and $b = 2$:
$x = -\frac{2}{2(1)} = -1$

To find the $y$-coordinate of the vertex, substitute $x = -1$ into the equation:
$y = (-1)^2 + 2(-1) + 3$ $y = 1 - 2 + 3 = 2$ Thus, the vertex is at $(-1, 2)$.

3. Find the y-intercept

The y-intercept occurs when $x = 0$:
$y = 0^2 + 2(0) + 3 = 3$ Thus, the y-intercept is $(0, 3)$.

4. Determine the axis of symmetry

The axis of symmetry is the vertical line that passes through the vertex. Its equation is:
$x = -1$

5. Identify additional points

To get more points for the graph, substitute nearby $x$-values:

• For $x = -2$:
  $y = (-2)^2 + 2(-2) + 3 = 4 - 4 + 3 = 3 \quad \text{(point: (-2, 3))}.$

• For $x = 1$:
  $y = (1)^2 + 2(1) + 3 = 1 + 2 + 3 = 6 \quad \text{(point: (1, 6))}.$

6. Sketch the graph

• The parabola opens upward.
• The vertex is at $(-1, 2)$.
• The y-intercept is $(0, 3)$.
• Symmetry helps reflect points across $x = -1$ to complete the graph.

Would you like me to plot this graph visually for you? Let me know! 😊

Related Questions:

1. How do you find the roots of a quadratic equation?
2. What happens to the graph if the coefficient $a$ changes?
3. How is the vertex form of a quadratic equation derived?
4. What are real-life applications of parabolas?
5. How do you determine if a quadratic has real or complex roots?

Tip: A quadratic graph is symmetric around its vertex—use this property to save time when sketching!