Consider the parabola given by the equation: f(x)=4x2+14x-14

Find the following for this parabola:

A) The vertex:   

B) The vertical intercept is the point   

C) Find the coordinates of the two x intercepts of the parabola and write them as a list, separated by commas:
  
It is OK to round your value(s) to to two decimal places.

To solve the given parabola equation \( f(x) = 4x^2 + 14x - 14 \), we'll follow the steps for each part of the question:

### A) The Vertex:
The vertex of a parabola in the form \( f(x) = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex:

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

For the given equation, \( a = 4 \), \( b = 14 \), and \( c = -14 \).

\[
x_{\text{vertex}} = -\frac{14}{2(4)} = -\frac{14}{8} = -1.75
\]

Now, substitute \( x = -1.75 \) into the original equation to find the y-coordinate of the vertex:

\[
f(-1.75) = 4(-1.75)^2 + 14(-1.75) - 14
\]
\[
f(-1.75) = 4(3.0625) + 14(-1.75) - 14
\]
\[
f(-1.75) = 12.25 - 24.5 - 14 = -26.25
\]

Thus, the vertex is:

\[
\boxed{(-1.75, -26.25)}
\]

### B) The Vertical Intercept:
The vertical intercept (also called the y-intercept) is the point where the parabola crosses the y-axis, which occurs when \( x = 0 \).

Substitute \( x = 0 \) into the equation:

\[
f(0) = 4(0)^2 + 14(0) - 14 = -14
\]

So, the vertical intercept is:

\[
\boxed{(0, -14)}
\]

### C) The x-Intercepts:
To find the x-intercepts, set \( f(x) = 0 \) and solve for \( x \):

\[
4x^2 + 14x - 14 = 0
\]

We can solve this using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = 14 \), and \( c = -14 \).

First, calculate the discriminant:

\[
\Delta = b^2 - 4ac = 14^2 - 4(4)(-14) = 196 + 224 = 420
\]

Now, apply the quadratic formula:

\[
x = \frac{-14 \pm \sqrt{420}}{2(4)} = \frac{-14 \pm \sqrt{420}}{8}
\]

The square root of 420 is approximately \( 20.49 \). Thus, we get two solutions:

\[
x_1 = \frac{-14 + 20.49}{8} = \frac{6.49}{8} \approx 0.81
\]
\[
x_2 = \frac{-14 - 20.49}{8} = \frac{-34.49}{8} \approx -4.31
\]

So, the x-intercepts are approximately:

\[
\boxed{(0.81, 0), (-4.31, 0)}
\]

### Summary:
- A) Vertex: \( (-1.75, -26.25) \)
- B) Vertical Intercept: \( (0, -14) \)
- C) X-Intercepts: \( 0.81, -4.31 \)

---

Would you like more details or have any questions?

Here are 5 follow-up questions to extend this topic:
1. How do you find the focus and directrix of a parabola?
2. Can you explain the role of the discriminant in determining the number of real solutions for x-intercepts?
3. What is the axis of symmetry for the parabola in this case?
4. How would the graph of the parabola change if the coefficient of \( x^2 \) were negative?
5. Can you derive the vertex form of the quadratic equation from its standard form?

**Tip**: The axis of symmetry of a parabola always passes through its vertex and has the equation \( x = \frac{-b}{2a} \).

Consider the parabola given by the equation: f(x)=4x^2+14x-14. Find the following for this parabola: A) The vertex. B) The vertical intercept is the point. C) Find the coordinates of the two x intercepts of the parabola and write them as a list, separated by commas: It is OK to round your value(s) to two decimal places.

Learn how to solve for the vertex, y-intercept, and x-intercepts of the parabola f(x)=4x^2+14x-14. This detailed solution covers vertex formula, solving quadratic equations, and using the quadratic formula to find x-intercepts. Ideal for students in Grades 9-10.

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