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The vertex of a parabola represents its highest or lowest point, depending on its orientation. In the context of quadratic functions, the vertex can be found by using the vertex formula, \(h = -\frac{b}{2a}\) and \(k = f(h)\). Here, \(h\) is the x-coordinate, and \(k\) is the y-coordinate of the vertex.
To find the vertex of a quadratic equation, make sure the equation is in standard form: \(f(x) = ax^2 + bx + c\). Once you have identified the values for \(a\) and \(b\), plug them into the formula for \(h\) to determine the x-coordinate. Then, substitute this value back into the equation to find the y-coordinate \(k\).
For the quadratic function \(f(x) = 6x^2 + 12x - 5\), we found the vertex was \((-1, -11)\). This tells us it is the lowest point on the graph because the parabola opens upwards (since \(a = 6\) is positive).

The x-intercepts of a quadratic equation are the points where the graph crosses the x-axis. These points occur when the value of the function is zero, i.e., \(f(x) = 0\).
To find the x-intercepts, use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(b^2 - 4ac\) is called the discriminant, and it determines whether there are real solutions (intercepts). A positive discriminant indicates two real x-intercepts, zero indicates a single intercept, and a negative discriminant means there are no real x-intercepts.
For \(f(x) = 6x^2 + 12x - 5\), substituting \(a = 6, b = 12, c = -5\) found the discriminant to be 264, resulting in the x-intercepts \(x \approx -0.097\) and \(x \approx -8.436\).

The y-intercept of a function is the point where the graph crosses the y-axis. It occurs when \(x = 0\). For any quadratic function given in standard form \(f(x) = ax^2 + bx + c\), finding the y-intercept is quite straightforward.
To determine the y-intercept, substitute \(x = 0\) into the function: \(f(0) = a(0)^2 + b(0) + c = c\). This means the y-intercept is always equal to the constant term \(c\).
In the function \(f(x) = 6x^2 + 12x - 5\), setting \(x = 0\) results in \(f(0) = -5\). Therefore, the y-intercept is at the point \((0, -5)\). This point is crucial when sketching the graph, providing a reference point for the shape of the parabola.

The quadratic formula is a powerful tool used to find the roots or x-intercepts of a quadratic equation. A quadratic equation typically takes the form \(ax^2 + bx + c = 0\).
The solution is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula always gives solutions for the x-values where the parabola intersects the x-axis. As mentioned earlier, the discriminant \(b^2 - 4ac\) within the formula gives insight into the nature of these solutions:

• Two real solutions when the discriminant is positive.
• One real solution when it equals zero.
• No real solutions when negative.

Applying the quadratic formula to \(f(x) = 6x^2 + 12x - 5\) provided the x-intercepts \(x \approx -0.097\) and \(x \approx -8.436\). This highlighted the characteristic two x-intercepts of this function with a positive discriminant.

A parabola is a U-shaped curve that can open either upwards or downwards. It is the graph of a quadratic function, represented by \(f(x) = ax^2 + bx + c\).
The direction in which a parabola opens correlates directly with the sign of \(a\):

• It opens upwards when \(a > 0\), making the vertex the lowest point.
• It opens downwards when \(a < 0\), and the vertex is the highest point.

The parabola has characteristic symmetry about its vertex. Knowing the vertex, x-intercepts, and y-intercept allows for sketching its graph.
For the function \(f(x) = 6x^2 + 12x - 5\), the parabola opens upwards because \(a = 6\) is positive. The graph includes the points where it intersects the axes, providing essential guidance for plotting the curve smoothly and accurately.