To solve this problem, we need to analyze the quadratic function:

$f(x) = 2x^2 + 4x + 3$

Here's a step-by-step solution to find the vertex, determine if the graph opens upward or downward, and identify any intercepts.

Step 1: Determine the Vertex

For a quadratic function in the form $f(x) = ax^2 + bx + c$, the vertex $(h, k)$ can be found using: $h = -\frac{b}{2a}$

In this case:

• $a = 2$
• $b = 4$
• $c = 3$

Substitute the values for $a$ and $b$: $h = -\frac{4}{2 \cdot 2} = -\frac{4}{4} = -1$

Now, find $k$ by substituting $x = -1$ back into the function: $k = f(-1) = 2(-1)^2 + 4(-1) + 3$ $k = 2 \cdot 1 - 4 + 3 = 2 - 4 + 3 = 1$

So, the vertex is $(-1, 1)$.

Step 2: Determine the Direction of the Parabola

The direction of the graph depends on the coefficient $a$:

• If $a > 0$, the graph opens upward.
• If $a < 0$, the graph opens downward.

Since $a = 2$, which is positive, the graph opens upward.

Step 3: Find the Intercepts

1. Y-intercept: Set $x = 0$. $f(0) = 2(0)^2 + 4(0) + 3 = 3$ So, the y-intercept is $(0, 3)$.

2. X-intercepts: Set $f(x) = 0$ and solve for $x$. $2x^2 + 4x + 3 = 0$ Divide by 2: $x^2 + 2x + \frac{3}{2} = 0$ This equation can be solved using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For $x^2 + 2x + \frac{3}{2} = 0$, $a = 1$, $b = 2$, and $c = \frac{3}{2}$: $x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot \frac{3}{2}}}{2 \cdot 1}$ $x = \frac{-2 \pm \sqrt{4 - 6}}{2}$ $x = \frac{-2 \pm \sqrt{-2}}{2}$ Since the discriminant is negative ($\sqrt{-2}$), there are no real x-intercepts.

Summary

• Vertex: $(-1, 1)$
• Opens: Upward
• Y-intercept: $(0, 3)$
• X-intercepts: None (no real solutions)

Would you like more details on any specific part, or do you have additional questions?

Further Questions:

1. How is the vertex form of a quadratic function derived from standard form?
2. How can you identify the axis of symmetry for a parabola from its equation?
3. What are the conditions under which a quadratic function has real x-intercepts?
4. How does the value of $a$ in $f(x) = ax^2 + bx + c$ affect the width of the parabola?
5. How do you complete the square to rewrite a quadratic function in vertex form?

Tip:

For any quadratic function $f(x) = ax^2 + bx + c$, use the formula $-\frac{b}{2a}$ to quickly find the x-coordinate of the vertex.