Question

Graph the quadratic y = x^2 + 2x + 4. What is the y-value of the y-value?

Answer 1

To graph the quadratic function \( y = x^2 + 2x + 4 \), we can start by identifying its key features:

### Step 1: Find the vertex
The vertex of a quadratic in the form \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex:

\[
x = -\frac{b}{2a}
\]

For our function \( y = x^2 + 2x + 4 \):
- \( a = 1 \)
- \( b = 2 \)

Substituting in the values:

\[
x = -\frac{2}{2 \cdot 1} = -1
\]

Next, we find the y-coordinate of the vertex by substituting \( x = -1 \) back into the function:

\[
y = (-1)^2 + 2(-1) + 4 = 1 - 2 + 4 = 3
\]

So, the vertex is at \( (-1, 3) \).

### Step 2: Find the y-intercept
The y-intercept occurs when \( x = 0 \):

\[
y = (0)^2 + 2(0) + 4 = 4
\]

Thus, the y-intercept is \( (0, 4) \).

### Step 3: Find the x-intercepts
To find the x-intercepts, we set \( y = 0 \):

\[
0 = x^2 + 2x + 4
\]

Now, we can use the quadratic formula where \( a = 1 \), \( b = 2 \), and \( c = 4 \):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}
\]
\[
x = \frac{-2 \pm \sqrt{4 - 16}}{2} = \frac{-2 \pm \sqrt{-12}}{2} = \frac{-2 \pm 2i\sqrt{3}}{2} = -1 \pm i\sqrt{3}
\]

Since the discriminant is negative, there are no real x-intercepts.

### Step 4: Sketch the graph
Using the vertex and y-intercept, we can sketch the graph:

- The vertex is at \( (-1, 3) \).
- The y-intercept is at \( (0, 4) \).
- The parabola opens upwards since \( a > 0 \).

### Conclusion
The vertex is the lowest point on the graph for this upward-opening parabola, and its y-coordinate is \( 3 \).

Thus, the y-value of the vertex is:

\[
\boxed{3}
\]