Question

Which is the graph of y equals x squared plus 4 x plus 3?
(1 point)
Responses

A parabola is shown on coordinate grid opening upward with vertex at left-parenthesis negative 2 comma negative 1 right-parenthesis, passing through left-parenthesis negative 3 comma 0 right-parenthesis.
Image with alt text: A parabola is shown on coordinate grid opening upward with vertex at left-parenthesis negative 2 comma negative 1 right-parenthesis, passing through left-parenthesis negative 3 comma 0 right-parenthesis.

A parabola is shown on coordinate grid opening upward with vertex at left-parenthesis 2 comma negative 1 right-parenthesis, passing through left-parenthesis 3 comma 0 right-parenthesis.
Image with alt text: A parabola is shown on coordinate grid opening upward with vertex at left-parenthesis 2 comma negative 1 right-parenthesis, passing through left-parenthesis 3 comma 0 right-parenthesis.

A parabola is shown on coordinate grid opening downward with vertex at left-parenthesis negative 1 comma 4 right-parenthesis, passing through left-parenthesis negative 3 comma 0 right-parenthesis.
Image with alt text: A parabola is shown on coordinate grid opening downward with vertex at left-parenthesis negative 1 comma 4 right-parenthesis, passing through left-parenthesis negative 3 comma 0 right-parenthesis.

A parabola is shown on coordinate grid opening downward with vertex at left-parenthesis 1 comma 4 right-parenthesis, passing through left-parenthesis 3 comma 0 right-parenthesis.
Image with alt text: A parabola is shown on coordinate grid opening downward with vertex at left-parenthesis 1 comma 4 right-parenthesis, passing through left-parenthesis 3 comma 0 right-parenthesis.
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Answer 1

To determine the correct graph of the equation \( y = x^2 + 4x + 3 \), we can analyze the function by completing the square or by finding the vertex.

### Step 1: Identify the vertex using the vertex formula
The vertex of a parabola in the form \( y = ax^2 + bx + c \) can be found using:
\[
x = -\frac{b}{2a}
\]
Here, \( a = 1 \) and \( b = 4 \):
\[
x = -\frac{4}{2 \cdot 1} = -2
\]

### Step 2: Calculate the y-coordinate
Now, substitute \( x = -2 \) back into the equation to find \( y \):
\[
y = (-2)^2 + 4(-2) + 3 = 4 - 8 + 3 = -1
\]
So, the vertex of the parabola is at \( (-2, -1) \).

### Step 3: Finding x-intercepts (roots)
To find the x-intercepts, we can set \( y = 0 \):
\[
0 = x^2 + 4x + 3
\]
This can be factored as:
\[
0 = (x + 1)(x + 3)
\]
Setting each factor to zero gives us:
\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]
\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]
So, the x-intercepts are at \( (-1, 0) \) and \( (-3, 0) \).

### Conclusion
Since the vertex is at \( (-2, -1) \) and the parabola opens upward (as the coefficient of \( x^2 \) is positive), the correct graph must show:
- A vertex at \( (-2, -1) \)
- The parabola passing through the x-intercept at \( (-3, 0) \)

Therefore, the correct response is the first option: **A parabola is shown on coordinate grid opening upward with vertex at \( (-2, -1) \), passing through \( (-3, 0) \)**.