91Ó°ÊÓ

The vertex of a parabola is a crucial point, revealing the parabola's turning point. It is the "peak" or "valley" of the graph, depending on whether the parabola opens upwards or downwards. For a quadratic function in standard form

• \( ax^2 + bx + c \),

we can find the vertex using a helpful formula. The formula for the x-coordinate of the vertex is given by \( x = -\frac{b}{2a} \). This helps pinpoint where the vertex lies on the x-axis.
For example, examining our quadratic function \(6x^2 + 12x - 5\), we have \(a = 6\) and \(b = 12\). By substituting these into the formula, the x-coordinate of the vertex becomes \(x = -\frac{12}{2 \times 6} = -1\). To find the y-coordinate, substitute \(x = -1\) back into the function:
\[ f(-1) = 6(-1)^2 + 12(-1) - 5 = -11 \]
Thus, the vertex of the parabola is \((-1, -11)\). This tells us that the graph of the parabola hits its lowest point at \((-1, -11)\), since the parabola opens upwards.

X-intercepts are the points where the graph of a function crosses the x-axis. These are the solutions to the equation when the output \(f(x)\) equals zero. Finding the x-intercepts of a quadratic function often involves solving a quadratic equation.
For the quadratic equation \(6x^2 + 12x - 5 = 0\), we apply the quadratic formula

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \),

to find these intercepts. Here, \(a = 6\), \(b = 12\), and \(c = -5\).
First, compute the discriminant \(b^2 - 4ac\), which aids in determining the number and type of solutions. Calculating this gives us \(144 + 120 = 264\). A positive discriminant indicates two real x-intercepts. Substitute these into the formula:
\[ x = \frac{-12 \pm \sqrt{264}}{12} \]
This simplifies to approximate x-intercepts \(x \approx 0.354\) and \(x \approx -2.354\). These points show where the parabola touches the x-axis.

The y-intercept is the point where the graph of the function crosses the y-axis. It is found by evaluating the function at \(x = 0\). For any quadratic function in the form

• \(ax^2 + bx + c\),

this is simply the constant term \(c\).
For our function, \(f(x) = 6x^2 + 12x - 5\), the y-intercept is calculated as:
\[ f(0) = 6(0)^2 + 12(0) - 5 = -5 \]
Thus, the graph of the quadratic functions meets the y-axis at the point \((0, -5)\). This indicates that when there is no x-value in the function, the output is

• \(-5\),

giving us the y-intercept.

The standard form of a quadratic equation is expressed as

• \(ax^2 + bx + c\),

where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). This form provides essential information about the quadratic, such as its direction, vertex, and intercepts.
In our example, the equation \(f(x) = 6x^2 + 12x - 5\) is already neatly arranged in standard form. Here:

• \(a = 6\)
• \(b = 12\)
• \(c = -5\)

The coefficient \(a\) dictates the direction and "width" of the parabola. Since \(a = 6\), a positive number, the parabola opens upwards. The larger the absolute value of \(a\), the steeper the parabola.
The constants \(b\) and \(c\) influence the graph's position and intercepts; \(b\) affects the vertex's location along the x-axis, while \(c\) directly indicates the y-intercept. Understanding these elements can greatly help in sketching and analyzing the characteristics of the parabola on a graph.