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We are given a quadratic function \( f(x) \), its vertex \((-3, 6)\), and one of its x-intercepts \(\left(-\frac{17}{4}, 0\right)\). We need to find the other x-intercept.

### Step 1: General form of a quadratic function
A quadratic function with vertex form is expressed as:

\[
f(x) = a(x - h)^2 + k
\]

where \((h, k)\) is the vertex. Substituting the vertex \((-3, 6)\):

\[
f(x) = a(x + 3)^2 + 6
\]

### Step 2: Use the given x-intercept
At the x-intercepts, \( f(x) = 0 \). We are given one x-intercept \( \left( -\frac{17}{4}, 0 \right) \), so we substitute \( x = -\frac{17}{4} \) and \( f(x) = 0 \) into the equation:

\[
0 = a\left( -\frac{17}{4} + 3 \right)^2 + 6
\]

Simplifying \( -\frac{17}{4} + 3 \):

\[
-\frac{17}{4} + 3 = -\frac{17}{4} + \frac{12}{4} = -\frac{5}{4}
\]

Substitute this into the equation:

\[
0 = a\left( -\frac{5}{4} \right)^2 + 6
\]
\[
0 = a\left( \frac{25}{16} \right) + 6
\]
\[
0 = \frac{25a}{16} + 6
\]

### Step 3: Solve for \(a\)

Solve for \(a\) by isolating it on one side:

\[
\frac{25a}{16} = -6
\]
\[
a = \frac{-6 \times 16}{25} = \frac{-96}{25}
\]

### Step 4: Write the equation of the quadratic function

Now that we know \(a = \frac{-96}{25}\), we substitute back into the vertex form equation:

\[
f(x) = \frac{-96}{25}(x + 3)^2 + 6
\]

### Step 5: Solve for the other x-intercept

The x-intercepts occur when \( f(x) = 0 \). Setting the equation to zero:

\[
0 = \frac{-96}{25}(x + 3)^2 + 6
\]

Move 6 to the other side:

\[
\frac{-96}{25}(x + 3)^2 = -6
\]

Divide both sides by \(\frac{-96}{25}\):

\[
(x + 3)^2 = \frac{-6 \times 25}{-96} = \frac{150}{96} = \frac{25}{16}
\]

Take the square root of both sides:

\[
x + 3 = \pm \frac{5}{4}
\]

Thus:

\[
x = -3 \pm \frac{5}{4}
\]

### Step 6: Find both intercepts

1. \( x = -3 + \frac{5}{4} = \frac{-12}{4} + \frac{5}{4} = \frac{-7}{4} \)
2. \( x = -3 - \frac{5}{4} = \frac{-12}{4} - \frac{5}{4} = \frac{-17}{4} \)

The other x-intercept is \( \left( \frac{-7}{4}, 0 \right) \).

### Answer:
The correct answer is \( \boxed{B} \), which is \( \left( -\frac{7}{4}, 0 \right) \).

---

Do you want further details or have any questions? Here are 5 related questions:

1. How is the vertex form of a quadratic equation derived?
2. Can a quadratic equation have two real x-intercepts, one real intercept, or none?
3. What is the significance of the value \(a\) in the vertex form of a quadratic equation?
4. How can you find the axis of symmetry of a quadratic function from its equation?
5. What does the discriminant of a quadratic equation tell you about the x-intercepts?

**Tip:** Always check both positive and negative roots when solving quadratic equations.

When the quadratic function f is graphed in the xy-plane, where y = f(x), its vertex is (-3, 6). One of the x-intercepts of this graph is (-17/4, 0). What is the other x-intercept of the graph?

Solve a quadratic function with vertex (-3, 6) and given one x-intercept (-17/4, 0) to find the other x-intercept. This problem involves the quadratic vertex form, x-intercepts, and symmetry of parabolas. Suitable for high school students in Grades 10-12.

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