Use the vertex and intercepts to sketch the graph of the quadratic function. Give the equation for the​ parabola's axis of symmetry.  Use the parabola to identify the​ function's domain and range.

​f(x)equals=8 x squared plus 16 x minus 78x2+16x−7

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Part 1

Use the graphing tool to graph the equation. Use the vertex and one of the intercepts when drawing the graph.

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We have the quadratic function \( f(x) = 8x^2 + 16x - 7 \). To graph this function and determine the vertex, intercepts, and axis of symmetry, let's break the process down step by step.

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

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

For the equation \( f(x) = 8x^2 + 16x - 7 \):
- \( a = 8 \)
- \( b = 16 \)

Using the formula:

\[
x_{\text{vertex}} = \frac{-16}{2(8)} = \frac{-16}{16} = -1
\]

Now substitute \( x = -1 \) into the function to find the \( y \)-coordinate of the vertex:

\[
f(-1) = 8(-1)^2 + 16(-1) - 7 = 8(1) - 16 - 7 = 8 - 16 - 7 = -15
\]

So, the vertex is \( (-1, -15) \).

### Step 2: Find the Axis of Symmetry
The axis of symmetry is the vertical line that passes through the vertex. For a quadratic function, the axis of symmetry is:

\[
x = x_{\text{vertex}}
\]

Thus, the axis of symmetry is:

\[
x = -1
\]

### Step 3: Find the y-Intercept
To find the \( y \)-intercept, set \( x = 0 \) in the equation:

\[
f(0) = 8(0)^2 + 16(0) - 7 = -7
\]

Thus, the \( y \)-intercept is \( (0, -7) \).

### Step 4: Find the x-Intercept(s)
To find the \( x \)-intercepts, set \( f(x) = 0 \) and solve for \( x \):

\[
8x^2 + 16x - 7 = 0
\]

We can solve this using the quadratic formula:

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

Substituting the values \( a = 8 \), \( b = 16 \), and \( c = -7 \):

\[
x = \frac{-16 \pm \sqrt{16^2 - 4(8)(-7)}}{2(8)}
\]

\[
x = \frac{-16 \pm \sqrt{256 + 224}}{16} = \frac{-16 \pm \sqrt{480}}{16}
\]

\[
x = \frac{-16 \pm 21.91}{16}
\]

Thus, the two solutions are:

\[
x = \frac{-16 + 21.91}{16} = 0.369 \quad \text{(approx.)}
\]

\[
x = \frac{-16 - 21.91}{16} = -2.37 \quad \text{(approx.)}
\]

So, the \( x \)-intercepts are approximately \( (0.369, 0) \) and \( (-2.37, 0) \).

### Step 5: Domain and Range
- **Domain**: The domain of any quadratic function is all real numbers, so the domain is:

\[
(-\infty, \infty)
\]

- **Range**: Since the parabola opens upwards (because \( a = 8 \) is positive), the range starts from the \( y \)-coordinate of the vertex and goes to infinity. Thus, the range is:

\[
[-15, \infty)
\]

### Summary:
- Vertex: \( (-1, -15) \)
- Axis of symmetry: \( x = -1 \)
- \( y \)-intercept: \( (0, -7) \)
- \( x \)-intercepts: \( (0.369, 0) \), \( (-2.37, 0) \)
- Domain: \( (-\infty, \infty) \)
- Range: \( [-15, \infty) \)

Would you like to explore the graph further or have any additional questions?

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Here are 5 related questions to expand your understanding:
1. How does the value of \( a \) affect the direction in which a parabola opens?
2. What is the significance of the discriminant in the quadratic formula for finding \( x \)-intercepts?
3. How can we convert a quadratic equation into vertex form?
4. How do changes in the \( c \)-value affect the graph of a quadratic function?
5. What are the real-world applications of quadratic functions?

**Tip:** Always double-check the axis of symmetry and vertex when graphing parabolas—they are key to accuracy!

Use the vertex and intercepts to sketch the graph of the quadratic function. Give the equation for the parabola's axis of symmetry. Use the parabola to identify the function's domain and range. f(x) = 8x^2 + 16x - 7

Learn how to graph the quadratic function f(x) = 8x^2 + 16x - 7 by identifying the vertex, axis of symmetry, and intercepts. This detailed guide also covers the function's domain and range, making it suitable for students in Grades 9-10.

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