91Ó°ÊÓ

A parabola is a symmetrical, curve-shaped graph that can either open upwards or downwards. This characteristic shape is a staple of quadratic equations, specifically those in the form of \( y = ax^2 + bx + c \). In our given equation, \( y = \frac{1}{3}x^2 \), notice the lack of \( bx \) and \( c \) terms, simplifying the curve to open upwards and align symmetrically along the y-axis.

• Positive \( a \) values result in a parabola that opens upwards.
• Negative \( a \) values would flip the parabola to open downwards.

The equation \( y = \frac{1}{3}x^2 \) specifically indicates a wide parabola, due to the factor \( \frac{1}{3} \) providing less steepness compared to \( y = x^2 \). This is because smaller absolute values of \( a \) (less than 1) cause a vertical compression, making the graph appear wider.

The \( x \)-intercepts of a graph are the points where the graph crosses the x-axis. To find them, we set \( y = 0 \) and solve for \( x \). With our particular equation, \( y = \frac{1}{3}x^{2} \), replacing \( y \) with 0 gives us:\[0 = \frac{1}{3}x^{2}\]Solving this equation reveals that \( x = 0 \). Thus, the graph crosses the x-axis at the origin \((0, 0)\). Having only one x-intercept is typical of parabolas that intersect the axis exactly at their vertex.
You can remember this step by observing that quadratic equations will often either:

• Have two \( x \)-intercepts on either side of the vertex when they open and intersect more distantly.
• Or, sometimes only touch the x-axis at the vertex, as in our equation.

A graph's \( y \)-intercept refers to where it crosses the y-axis. This is found by setting \( x = 0 \) in the equation. For the equation \( y = \frac{1}{3}x^{2} \), substituting \( x \) with zero gives:\[y = \frac{1}{3}(0)^{2} = 0\]So, the \( y \)-intercept also lies at the origin \((0, 0)\). Interestingly, this point is the same for both the x- and y-intercepts in this case, implying that our parabola begins from this point.
It's important to recognize that for most parabolas, unless shifted, the y-intercept is simply the value of \( c \) in \( y = ax^2 + bx + c \). In our case, \( c = 0 \), hence the intercept at the origin.

The vertex of a parabola is a significant point since it is the point of maximum or minimum value of the graph, and where it changes direction. For a parabola given in the form \( y = ax^2 + bx + c \), the vertex coordinates can be deduced using the formula:\[x = -\frac{b}{2a}\]Since our equation \( y = \frac{1}{3}x^{2} \) lacks \( b \) and \( c \) terms, the vertex is straightforwardly at \((0, 0)\). This equation showcases the simplest scenario where the vertex doubles as both the x- and y-intercepts.
Interestingly, due to the vertex at the origin, you can see that the vertex also indicates the axis of symmetry, which in this equation is aligned with the y-axis.

Graph sketching involves drawing the basic shape of the quadratic equation on a coordinate plane. For \( y = \frac{1}{3}x^2 \), the key is understanding its features:

• Vertex at \((0, 0)\).
• Opening upwards.
• Wider than standard parabola due to \( a = \frac{1}{3} \).

Begin by plotting the intercepts. Here, both x- and y-intercepts are at \((0, 0)\). As there are no extreme points to mark beyond the vertex itself, focus on the curvature.
You can achieve this by plotting a few x-values to see how wide the parabola indeed spreads, such as substituting \( x = 1 \) and \( x = -1 \) to show points like \((1, \frac{1}{3})\) and \((-1, \frac{1}{3})\). These points, when connected smoothly through the vertex, illustrate the full, visually broad shape of this quadratic curve. Finally, always ensure symmetry across the y-axis to complete your sketch.