Let's solve the parts of the problem one by one for the given equation $f(x) = -4x^2 + 14x + 11$.

A) Finding the Vertex

The vertex of a parabola given by $f(x) = ax^2 + bx + c$ can be found using the formula for the $x$-coordinate of the vertex: $x = -\frac{b}{2a}$ In this case, $a = -4$ and $b = 14$.

1. Calculate the $x$-coordinate of the vertex: $x = -\frac{14}{2 \times -4} = \frac{14}{8} = 1.75$

2. Substitute $x = 1.75$ back into the equation to find the $y$-coordinate of the vertex: $f(1.75) = -4(1.75)^2 + 14(1.75) + 11$ $= -4(3.0625) + 24.5 + 11$ $= -12.25 + 24.5 + 11 = 23.25$

So, the vertex is at: $(1.75, 23.25)$

B) Finding the Vertical Intercept

The vertical intercept occurs when $x = 0$. Substitute $x = 0$ into the equation:

$f(0) = -4(0)^2 + 14(0) + 11 = 11$

Thus, the vertical intercept is: $(0, 11)$

C) Finding the $x$-Intercepts

To find the $x$-intercepts, we set $f(x) = 0$:

$-4x^2 + 14x + 11 = 0$

This is a quadratic equation in the form $ax^2 + bx + c = 0$. We can solve for $x$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our equation:

• $a = -4$
• $b = 14$
• $c = 11$

1. Calculate the discriminant: $b^2 - 4ac = 14^2 - 4(-4)(11) = 196 + 176 = 372$

2. Substitute into the quadratic formula: $x = \frac{-14 \pm \sqrt{372}}{2 \times -4}$ $x = \frac{-14 \pm 19.29}{-8}$

3. Solve for the two values of $x$:

   • For $x_1$: $x_1 = \frac{-14 + 19.29}{-8} = \frac{5.29}{-8} = -0.66$

   • For $x_2$: $x_2 = \frac{-14 - 19.29}{-8} = \frac{-33.29}{-8} = 4.16$

The $x$-intercepts are approximately: $(-0.66, 0) \text{ and } (4.16, 0)$

Summary of Results:

• A) Vertex: $(1.75, 23.25)$
• B) Vertical Intercept: $(0, 11)$
• C) $x$-Intercepts: $(-0.66, 0), (4.16, 0)$

Would you like further details on any part of this solution?

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Here are some related questions to deepen your understanding:

1. How do we derive the formula for the vertex of a parabola?
2. Why does the sign of $a$ determine the direction in which the parabola opens?
3. Can you explain the role of the discriminant in finding real or complex solutions?
4. What is the significance of the vertex in real-world applications of parabolas?
5. How does changing the values of $a$, $b$, and $c$ affect the shape and position of the parabola?

Tip: Remember that the $x$-coordinate of the vertex is always at $x = -\frac{b}{2a}$ for any quadratic equation.