Question

1. which is the graph of $f(x) = x^2 + 4x + 2$?

Explanation:

Step1: Determine the parabola's direction

The function \( f(x) = x^2 + 4x + 2 \) is a quadratic function in the form \( f(x)=ax^2+bx+c \), where \( a = 1 \), \( b = 4 \), and \( c = 2 \). Since \( a=1>0 \), the parabola opens upwards.

Step2: Find the vertex's x - coordinate

The formula for the x - coordinate of the vertex of a quadratic function \( f(x)=ax^2 + bx + c \) is \( x=-\frac{b}{2a} \). Substituting \( a = 1 \) and \( b = 4 \) into the formula, we get \( x=-\frac{4}{2\times1}=- 2 \).

Step3: Find the vertex's y - coordinate

Substitute \( x=-2 \) into the function \( f(x)=x^2 + 4x + 2 \). Then \( f(-2)=(-2)^2+4\times(-2)+2=4 - 8 + 2=-2 \). So the vertex of the parabola is at \( (-2,-2) \).

Step4: Analyze the y - intercept

To find the y - intercept, we set \( x = 0 \). Then \( f(0)=0^2+4\times0 + 2=2 \). So the parabola intersects the y - axis at \( (0,2) \).

We can also rewrite the function in vertex form. Completing the square for \( f(x)=x^2 + 4x + 2 \):

\( f(x)=x^2+4x + 4-4 + 2=(x + 2)^2-2 \)

This confirms the vertex is \( (-2,-2) \) and the parabola opens upwards with a y - intercept at \( (0,2) \). When looking at the graphs, we should look for a parabola that opens upwards, has its vertex at \( (-2,-2) \) and passes through \( (0,2) \).

Answer:

The graph of \( f(x)=x^2 + 4x + 2 \) is a parabola opening upwards with vertex at \((-2,-2)\) and y - intercept at \((0,2)\). (To identify the specific graph among given options, look for the parabola with these characteristics: opening up, vertex at \((-2,-2)\), passing through \((0,2)\))