Describe the graph of the quadratic function.
f(x) = x2 + 10x + 7
 parabola opening upward
parabola opening downward    
Correct: Your answer is correct.

Identify the vertex and the x-intercept(s). Use a graphing utility to verify your results. (If an answer does not exist, enter DNE.)
vertex    	
(x, y) =

 
  
Correct: Your answer is correct.
 

x-intercept    	
(x, y) =

 
  
Incorrect: Your answer is incorrect.
 

    (smaller x-value)
x-intercept    	
(x, y) =

 
  
Incorrect: Your answer is incorrect.
 

    (larger x-value)


The quadratic function is given as:

\[
f(x) = x^2 + 10x + 7
\]

### Step 1: Identify the vertex
The vertex of a parabola in the form \( f(x) = ax^2 + bx + c \) can be found using the formula:

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

For the given function \( f(x) = x^2 + 10x + 7 \), we have \( a = 1 \) and \( b = 10 \). Substituting into the formula:

\[
x = -\frac{10}{2(1)} = -5
\]

To find the corresponding \( y \)-coordinate of the vertex, substitute \( x = -5 \) back into the function:

\[
f(-5) = (-5)^2 + 10(-5) + 7 = 25 - 50 + 7 = -18
\]

Thus, the vertex is:

\[
(x, y) = (-5, -18)
\]

### Step 2: Find the x-intercepts
The x-intercepts occur where \( f(x) = 0 \). To find them, we need to solve the quadratic equation:

\[
x^2 + 10x + 7 = 0
\]

We can solve this using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For \( a = 1 \), \( b = 10 \), and \( c = 7 \):

\[
x = \frac{-10 \pm \sqrt{10^2 - 4(1)(7)}}{2(1)} = \frac{-10 \pm \sqrt{100 - 28}}{2} = \frac{-10 \pm \sqrt{72}}{2} = \frac{-10 \pm 6\sqrt{2}}{2}
\]

Thus, the two solutions are:

\[
x = \frac{-10 + 6\sqrt{2}}{2} \quad \text{and} \quad x = \frac{-10 - 6\sqrt{2}}{2}
\]

Simplifying these expressions:

\[
x_1 = -5 + 3\sqrt{2} \quad \text{and} \quad x_2 = -5 - 3\sqrt{2}
\]

Numerically, these are approximately:

\[
x_1 \approx -0.757 \quad \text{and} \quad x_2 \approx -9.243
\]

Therefore, the x-intercepts are:

\[
(x_1, 0) \approx (-0.757, 0) \quad \text{and} \quad (x_2, 0) \approx (-9.243, 0)
\]

### Step 3: Graphing Utility Verification
You can verify these results by using a graphing utility (e.g., Desmos or a graphing calculator). The graph of the parabola will open upwards, have its vertex at \( (-5, -18) \), and cross the x-axis at \( (-0.757, 0) \) and \( (-9.243, 0) \).

---

### Summary:
- **Vertex**: \( (-5, -18) \)
- **X-intercepts**: \( (-0.757, 0) \) and \( (-9.243, 0) \)

Do you need more details or further clarification?

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Here are 5 related questions:
1. How do you find the y-intercept of a quadratic function?
2. What effect does the coefficient \( a \) have on the direction of a parabola?
3. How do you convert a quadratic function from standard form to vertex form?
4. What is the discriminant, and how does it determine the number of x-intercepts?
5. Can a parabola have no x-intercepts? If so, when?

Tip: The vertex form of a quadratic function is very useful for identifying the vertex directly!

Describe the graph of the quadratic function. f(x) = x^2 + 10x + 7. Identify the vertex and the x-intercept(s). Use a graphing utility to verify your results.

This page provides a detailed solution to graphing the quadratic function f(x) = x^2 + 10x + 7. It includes the vertex calculation, x-intercepts using the quadratic formula, and graphical verification. Ideal for Grades 9-11.

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