Verified Answer :

  • The parabola \(y = -3x^2 + 12x - 4\) opens downward because the leading coefficient is negative.
  • The vertex is found at \((2, 8)\) using the formula \(x_v = -\frac{b}{2a}\).
  • The y-intercept is at \((0, -4)\), and the x-intercepts are approximately \(0.367\) and \(3.633\).
  • The axis of symmetry is the line \(x = 2\).
  • The vertex is \(\boxed{(2, 8)}\).

Explanation

  1. Analyzing the Quadratic Function

The quadratic function is given by \(y = -3x^2 + 12x - 4\). This is a downward-opening parabola because the leading coefficient \(a = -3\) is negative. To understand the geometry of this curve, we identify several key features: the vertex, the y-intercept, the x-intercepts, and the axis of symmetry.

  1. Finding the Vertex and Axis of Symmetry

The vertex of a parabola \(y = ax^2 + bx + c\) occurs at \(x_v = -\frac{b}{2a}\). Substituting the coefficients \(a = -3\) and \(b = 12\), we get \(x_v = -\frac{12}{2(-3)} = \frac{-12}{-6} = 2\). The y-coordinate is found by evaluating the function at \(x_v = 2\): \(y_v = -3(2)^2 + 12(2) - 4 = -3(4) + 24 - 4 = -12 + 24 - 4 = 8\). Thus, the vertex is \((2, 8)\). The axis of symmetry is the vertical line passing through the vertex, given by \(x = 2\).

  1. Identifying Intercepts

The y-intercept is found by evaluating the function at \(x = 0\): \(y = -3(0)^2 + 12(0) - 4 = -4\). This gives the point \((0, -4)\). To find the x-intercepts, we solve \(-3x^2 + 12x - 4 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
Substituting the values, we have:
\[x = \frac{-12 \pm \sqrt{12^2 - 4(-3)(-4)}}{2(-3)} = \frac{-12 \pm \sqrt{144 - 48}}{-6} = \frac{-12 \pm \sqrt{96}}{-6} = \frac{-12 \pm 4\sqrt{6}}{-6} = 2 \mp \frac{2\sqrt{6}}{3}\]
Using the approximation \(\sqrt{6} \approx 2.449\), the intercepts are approximately:
\(x_1 = 2 - \frac{2(2.449)}{3} \approx 2 - 1.633 = 0.367\)
\(x_2 = 2 + \frac{2(2.449)}{3} \approx 2 + 1.633 = 3.633\)
The approximate coordinates are \((0.367, 0)\) and \((3.633, 0)\).

  1. Graphing the Parabola

To sketch the parabola:

  1. Plot the vertex at \((2, 8)\).
  2. Plot the y-intercept at \((0, -4)\). By symmetry, the point \((4, -4)\) must also be on the parabola since it is 2 units to the right of the axis of symmetry.
  3. Plot the approximate x-intercepts at \((0.367, 0)\) and \((3.633, 0)\).
  4. Draw a smooth, symmetric curve connecting these points, opening downwards. The final graph represents the parabola \(y = -3x^2 + 12x - 4\). The coordinates are shown as \(\boxed{(2, 8)}\) for the vertex.

Examples

Quadratic functions are widely used to model trajectories in physics. For example, if you throw a ball into the air, its height over time follows a parabolic path. The vertex represents the peak height the ball reaches, and the x-intercepts show when the ball was thrown and when it hits the ground. Understanding how to find these points allows you to predict the behavior of moving objects, design bridges, or even optimize profit in business models!

Related Questions