91Ó°ÊÓ

The vertex of a parabola is a crucial point that represents the peak or the lowest tip of a parabola, depending on its orientation. In the context of the quadratic function, the vertex helps determine whether the parabola has a maximum or a minimum value. In our example, the quadratic equation is \(y = x^2 + 6x - 1\). To find the vertex, we use the vertex formula for a quadratic equation in standard form \(y = ax^2 + bx + c\):

\[x = -\frac{b}{2a}\]

Substituting the values from our equation where \(a = 1\) and \(b = 6\), we arrive at:

• \(x = -\frac{6}{2 \cdot 1} = -3\)

Next, substitute \(x = -3\) back into the original equation to find \(y\):

• \(y = (-3)^2 + 6(-3) - 1 = -10\)

The vertex is \((-3, -10)\), signifying the lowest point of the parabola as it opens upwards.

The axis of symmetry of a parabola is an imaginary vertical line that passes through the vertex, dividing the parabola into two mirror-image halves. This axis provides a valuable insight into the symmetry of the quadratic function. The equation for the axis of symmetry can be represented in terms of the x-coordinate of the vertex.

For the quadratic equation \(y = ax^2 + bx + c\), the axis of symmetry formula is:

• \(x = -\frac{b}{2a}\)

In our example's quadratic equation, substituting \(b = 6\) and \(a = 1\), we computed:

• \(x = -\frac{6}{2 \cdot 1} = -3\)

Thus, the axis of symmetry is the vertical line \(x = -3\). This line is crucial to understand because the parabola is perfectly mirrored around this line, ensuring that the values on either side of this axis are equidistant to it.

The x-intercepts are points where the graph of the polynomial crosses the x-axis, meaning the y-value is zero at these points. Finding x-intercepts involves solving \(0 = ax^2 + bx + c\). These intercepts are the roots of the quadratic equation and can be found using the quadratic formula:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

For our function \(y = x^2 + 6x - 1\):

• \(a = 1\)
• \(b = 6\)
• \(c = -1\)

Substituting these into the quadratic formula gives:

• \(x = \frac{-6 \pm \sqrt{36 + 4}}{2}\)
• Simplifying further gives \(x = \frac{-6 \pm 2\sqrt{10}}{2} = -3 \pm \sqrt{10}\)

This results in two x-intercepts: \((-3 + \sqrt{10}, 0)\) and \((-3 - \sqrt{10}, 0)\). These points are where the parabola meets the x-axis.

The y-intercept of a quadratic function is the point where the graph of the function crosses the y-axis. This happens where \(x = 0\). To find the y-intercept of a quadratic equation \(y = ax^2 + bx + c\), you simply substitute \(x = 0\) into the equation.

For our equation \(y = x^2 + 6x - 1\), substituting \(x = 0\) gives us:

• \(y = 0^2 + 6 \cdot 0 - 1 = -1\)

The y-intercept is therefore \((0, -1)\). This point is essential for graphing the function as it provides a fixed reference point from which the curve will extend. By understanding how the graph intersects the y-axis, you get further insight into the parabola's behavior.