91Ó°ÊÓ

A quadratic function is a type of polynomial that is represented by the equation \( ax^2 + bx + c \). This is known as the "standard form" of a quadratic function. The standard form is essential as it allows you to identify the coefficients \( a \), \( b \), and \( c \), which are crucial for further calculations, like finding the vertex or determining the direction of the parabola.
The given function \( f(x) = x + x^2 \) can look a little different at first, but by rearranging the terms, it becomes \( f(x) = x^2 + x \). Here, the standard form is achieved with \( a = 1 \), \( b = 1 \), and \( c = 0 \). This form neatly organizes the function, making it easy to recognize and work with. Analyzing a quadratic function in standard form is the foundational step to effectively sketch and understand its graph.

Some key takeaways from standard form include:

• Locating coefficients \( a \), \( b \), and \( c \)
• Understanding the direction of the parabola—upwards if \( a > 0 \) and downwards if \( a < 0 \)
• Ready setup for other calculations like finding the vertex or intercepts

Finding the vertex of a quadratic function is important because it represents the highest or lowest point on its graph (depending on whether the parabola opens upward or downward). The vertex formula, \( x = -\frac{b}{2a} \), helps in finding the x-coordinate of the vertex. Once we have the x-coordinate, we plug it back into the function to find the y-coordinate.

For the quadratic function \( f(x) = x^2 + x \), we substitute \( a = 1 \) and \( b = 1 \) into the vertex formula:\[ x = -\frac{1}{2 \cdot 1} = -\frac{1}{2} \] Now, to find the y-coordinate, we substitute \( x = -\frac{1}{2} \) back into the function: \[ f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \] Thus, the vertex is at \( \left(-\frac{1}{2}, -\frac{1}{4}\right) \). Given that \( a > 0 \) (since \( a = 1 \)), the parabola opens upwards, signifying that this vertex is a minimum point of the function.

Key points about the vertex:

• The vertex gives the maximum or minimum value of the function
• The x-coordinate is found using \(-\frac{b}{2a}\)
• Offers a pivotal point for sketching the parabola

Sketching a parabola from a quadratic function provides visual insight into its properties and behavior. Given the function \( f(x) = x^2 + x \), sketching starts with identifying key features like the vertex and the y-intercept.

To start the sketch:

• Plot the vertex at \( \left(-\frac{1}{2}, -\frac{1}{4}\right) \)
• Determine the y-intercept, \( f(0) = 0 \). This is another crucial point since it shows where the parabola crosses the y-axis.
• Choose additional points for accuracy; in this case, calculating \( f(1) = 2 \) provides another point \( (1, 2) \).

More points make the sketch more precise and demonstrate the symmetric nature of parabolas. Here, as \( a = 1 \), the parabola gushes upwards from the minimum point at the vertex. This uplifting trend often signifies a promising future for positive "x" scenarios!

Remember, for an accurate parabola sketch:

• Start with the vertex and intercepts
• Use symmetry, positive means it opens upwards
• Plotting additional points deepens insight into its curvature