y=3(x+1)^2-7

The equation you've provided is in vertex form, which is used to represent a quadratic function. The general form of a quadratic function in vertex form is:

\[ y = a(x-h)^2 + k \]

Where:

• \((h, k)\) is the vertex of the parabola,
• \(a\) determines the direction and width of the parabola.

For the equation \( y = 3(x + 1)^2 - 7 \):

• The value of \(a\) is \(3\), meaning the parabola opens upwards (since \(a > 0\)).
• The vertex \((h, k)\) can be identified from the equation as follows:
  • \(h = -1\) (since \(x + 1\) translates to \(x - (-1)\))
  • \(k = -7\)

Thus, the vertex of the parabola is \((-1, -7)\).

Here are some key features of this parabola based on your equation:

1. Direction: Opens upwards.

2. Vertex: \((-1, -7)\).

3. Axis of symmetry: The line \(x = -1\).

4. Y-intercept: To find the y-intercept, set \(x = 0\): \[ y = 3(0 + 1)^2 - 7 = 3(1) - 7 = 3 - 7 = -4 \] So, the y-intercept is \((0, -4)\).

5. X-intercepts: To find the x-intercepts, set \(y = 0\): \[ 0 = 3(x + 1)^2 - 7 \] \[ 3(x + 1)^2 = 7 \] \[ (x + 1)^2 = \frac{7}{3} \] Taking the square root of both sides gives: \[ x + 1 = \pm \sqrt{\frac{7}{3}} \] Therefore: \[ x = -1 \pm \sqrt{\frac{7}{3}} \]

6. Graph: The parabola opens upwards with the vertex at \((-1, -7)\) and has the points calculated above.

If you need another specific aspect or further calculations, feel free to ask!