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: $0.81, -4.31$

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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 $x = \frac{-b}{2a}$.