91Ó°ÊÓ

The standard form of a quadratic function is expressed as \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. These constants dictate different characteristics of the parabola, such as its width, direction, and position. For our given function, \( f(x) = 6x^2 + 12x - 5 \), we can easily see that it is already in standard form because it fits the pattern.

• \( a = 6 \)
• \( b = 12 \)
• \( c = -5 \)

These values will not only help us in graphing the parabola but also in finding other important features like the vertex and intercepts. The standard form is a crucial starting point for analyzing quadratic functions because it gives a clear path to these key components.

The vertex of a quadratic function is a crucial point that represents either the maximum or minimum value of the function, depending on the direction the parabola opens. For a function in standard form \( ax^2 + bx + c \), the \( x \)-coordinate of the vertex is calculated using the formula \( x = -\frac{b}{2a} \).
In our example, substituting \( a = 6 \) and \( b = 12 \), we find \( x = -\frac{12}{12} = -1 \). The \( y \)-coordinate, often denoted as \( k \), is found by substituting \( x = -1 \) back into the function: \( f(-1) = 6(-1)^2 + 12(-1) - 5 = -11 \).
Therefore, the vertex of this quadratic function is \((-1, -11)\), showing it is the lowest point because the parabola opens upwards (since \( a > 0 \)). It's the turning point of the function and gives symmetry to the parabola.

The x-intercepts of a quadratic function are where the graph crosses the x-axis, and they can be found by setting the function \( f(x) = ax^2 + bx + c \) equal to zero and solving for \( x \). This can be done using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our function, substituting the values \( a = 6 \), \( b = 12 \), and \( c = -5 \) gives: \[ x = \frac{-12 \pm \sqrt{144 + 120}}{12} = \frac{-12 \pm \sqrt{264}}{12} \] By simplifying, the approximate intercepts are about \( x \approx -0.31 \) and \( x \approx -2.69 \).
These roots are also known as the solutions to the equation and give important information about the function's behavior.

The y-intercept of a quadratic function indicates where the graph crosses the y-axis. To find this, we evaluate the function \( f(x) = ax^2 + bx + c \) at \( x = 0 \).
For our function \( f(x) = 6x^2 + 12x - 5 \), the calculation is straightforward: \[ f(0) = 6(0)^2 + 12(0) - 5 = -5 \] This means the y-intercept is at the point \((0, -5)\). The y-intercept provides a starting point for graphing because it indicates the initial value of the function when no other variable is considered.
Keep in mind that unless \( b = 0 \), the y-intercept is unique since the quadratic term and the linear term become zero when \( x = 0 \). This makes the y-intercept the constant term \( c \).