91Ó°ÊÓ

To understand the vertex of a parabola, it's helpful to think of it as the "turning point" where the curve changes direction. For quadratic functions in the form of \( ax^2 + bx + c \), the vertex can be calculated using the formula \( x = -\frac{b}{2a} \). This formula helps you determine the \( x \)-coordinate of the vertex.
Once you find this value, you plug it back into the original equation to find the \( y \)-coordinate.For the quadratic function \( f(x) = -x^2 - 4x + 4 \), the values of \( a \) and \( b \) are \( -1 \) and \( -4 \) respectively. Using the vertex formula, we calculate:

• \( x = -\frac{-4}{2(-1)} = 2 \)
• \( f(2) = -(2)^2 - 4(2) + 4 = -8 \)

The vertex of the parabola is therefore at the point \((2, -8)\). This point indicates that the parabola reaches its maximum downward point here since the \( a \) value is negative.

The \( x \)-intercepts of a quadratic function are the points at which the parabola crosses the \( x \)-axis, meaning the \( y \)-value at these points is zero. To find these, you set the quadratic equation equal to zero and solve for \( x \).For our function \( -x^2 - 4x + 4 \), the equation becomes:

• \( -x^2 - 4x + 4 = 0 \)

One way to solve this is to use the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• With \( a = -1, b = -4, \) and \( c = 4 \)
  • \( x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(-1)(4)}}{2(-1)} \)
  • \( x = \frac{4 \pm \sqrt{32}}{-2} \)
  After simplification, the intercepts are \( x = -2 \pm 2\sqrt{2} \).

This results in two \( x \)-intercepts, which are the points \((-2 + 2\sqrt{2}, 0)\) and \((-2 - 2\sqrt{2}, 0)\).

The \( y \)-intercept of a quadratic function is the point where the graph crosses the \( y \)-axis. This is easily found by setting \( x = 0 \) in the quadratic equation and solving for \( f(x) \).In our function \( f(x) = -x^2 - 4x + 4 \), substituting \( x = 0 \) gives:

• \( f(0) = -(0)^2 - 4(0) + 4 = 4 \)

Therefore, the \( y \)-intercept is at the point \((0, 4)\).This is the exact spot where the graph meets the \( y \)-axis and helps give another valuable point to plot the quadratic graph.

Graphing a quadratic equation like \( f(x) = -x^2 - 4x + 4 \) involves plotting several key points and understanding the shape of a parabola. Quadratic functions graph as U-shaped curves, and knowing whether they open up or down is crucial.Here's how you can sketch the graph:

• Plot the vertex at \((2, -8)\). This point is central for drawing the parabola's shape.
• Identify the \( x \)-intercepts at \((-2 + 2\sqrt{2}, 0)\) and \((-2 - 2\sqrt{2}, 0)\).
• Mark the \( y \)-intercept at \((0, 4)\).
• Because the coefficient \( a = -1 \) is negative, the parabola opens downward, forming an upside-down U.

Connect these points smoothly. The vertex is the highest point due to the downward opening. This visualization helps you understand the real-world applications of quadratic functions, such as projectile motion or optimization problems.