The given quadratic function is:

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

Let's solve part (a) step by step.

1. Find the Vertex:

The vertex of a quadratic function in the form $f(x) = ax^2 + bx + c$ is found using the formula for the x-coordinate of the vertex:

$x_{\text{vertex}} = \frac{-b}{2a}$

For the given function $f(x) = x^2 + 2x$, we have:

• $a = 1$
• $b = 2$

Using the formula for $x_{\text{vertex}}$:

$x_{\text{vertex}} = \frac{-2}{2(1)} = -1$

Now, to find the y-coordinate of the vertex, we substitute $x = -1$ into the function:

$y_{\text{vertex}} = f(-1) = (-1)^2 + 2(-1) = 1 - 2 = -1$

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

2. Axis of Symmetry:

The axis of symmetry is the vertical line that passes through the x-coordinate of the vertex. So, the axis of symmetry is:

$x = -1$

3. Concavity:

The concavity of a quadratic function is determined by the sign of the coefficient $a$. In this case, $a = 1$, which is positive. Therefore, the parabola opens upward and is concave up.

Summary for Part (a):

• The vertex is $(-1, -1)$.
• The axis of symmetry is $x = -1$.
• The graph is concave up.

Would you like to explore further details or have any questions about this?

Additional Questions:

1. What is the significance of the vertex in a quadratic function?
2. How can you graph a quadratic function from its vertex form?
3. What changes in the equation $f(x) = ax^2 + bx + c$ affect the width and direction of the parabola?
4. How does the axis of symmetry help in graphing quadratic functions?
5. Can you find the x-intercepts and y-intercepts of the quadratic function?

Tip:

When analyzing the graph of a quadratic function, always start by determining the vertex and axis of symmetry. These give you a foundation for understanding the function's shape and behavior.