Problem 20 - College Algebra and Calculus: An Applied Approach

Question

Sketch the graph of the quadratic function. Identify the vertex and intercepts. $$f(x)=-2 x^{2}$$

Step-by-Step Solution

Answer

The vertex of the function is at the point (0,0). The intercepts of the function are also at (0,0). The graph of the function is a downward-facing parabola that passes through the origin.

Step 1: Identify the vertex

For a quadratic function in the form \(f(x) = ax^2 + bx + c\), the vertex is given by \((-b/2a , f(-b/2a))\). Our function is \(f(x) = -2x^2\), which can also be written as \(f(x) = -2x^2 + 0x + 0\). Hence, \(a = -2, b = 0, c = 0\). Plugging these into the vertex formula, the vertex will be \((-b/2a , f(-b/2a)) = (0 , f(0)) = (0 , 0)\). So, the vertex is at the origin.

Step 2: Find the intercepts

The x-intercept(s) of a function happen where \(f(x) = 0\). So set \(f(x) = -2x^2\) equal to zero and solve for x. Doing so, we get \(0 = -2x^2\), which simplifies to \(x = 0\). So, the x-intercept is at \(x = 0\). The y-intercept of a function is found by finding \(f(0)\). For this function, \(f(0) = -2(0)^2 = 0\). So the y-intercept is at \(y = 0\). Since x-intercept and y-intercept are same, the graph touches the origin.

Step 3: Sketch the graph

Knowing the vertex is at the origin (0,0) and the intercepts too are at the origin, plot these points on a graph. Because \(a = -2\) is negative, the graph of our function will open downwards. Draw a parabola that passes through these points and opens downwards.

Key Concepts

Vertex

In a quadratic function, the vertex is like the "tip" or the "turning point" of the parabola. It is crucial for understanding the overall shape and position of the graph. For the function given, which is in the form \(f(x) = ax^2 + bx + c\), the vertex formula is \((-\frac{b}{2a} , f(-\frac{b}{2a}))\). This formula helps you find where the vertex lies on the graph.

Consider the function \(f(x) = -2x^2\). Here, \(a = -2\), \(b = 0\), and \(c = 0\). By substituting into the vertex formula, we get:

• \(-\frac{b}{2a} = -\frac{0}{2(-2)} = 0\)
• Plugging \(x = 0\) back into the function gives \(f(0) = 0\).
• Thus, the vertex is \((0, 0)\).

This vertex is located at the origin, which means it is the central point where the parabola changes direction. Being at the origin implies symmetry around this point.

Intercepts

Intercepts are points where the graph meets the axes, providing vital clues to the graph's behavior.

First, we consider the **x-intercept(s)**. These occur where the function value equals zero, i.e., \(f(x) = 0\). For our function, this means: \(-2x^2 = 0\). Solving gives:

• \(x = 0\)

There is only one x-intercept, located at the origin (\(0,0\)).

Next, the **y-intercept** is found by evaluating the function at \(x = 0\), which here also happens to be \(f(0) = 0\).

• So, the y-intercept is also at \(y = 0\).

For this particular function, both intercepts coincide at the origin, making (0,0) a point where both axes are touched simultaneously by the graph.

Graphing Parabolas

Graphing parabolas involves utilizing both the vertex and intercepts to lay out a clear picture of the parabola on the graph.

Since the function in question is \(f(x) = -2x^2\), and we have discovered both the vertex and intercepts: the vertex at (0,0) and both intercepts also at (0,0), the graph starts to take shape inward.

• Given \(a = -2\) is negative, the graph opens downward.
• The steepness of the parabola is influenced by the value of \(a\). A more negative \(a\) indicates a steeper "U" shape.

When you plot these pieces of information, you start at the origin (touchpoint of both vertex and intercepts) and draw a symmetrical parabolic curve that dips downward.

Visualizing this helps in seeing how quickly the graph diverges away from the vertex, widening as it descends further into negative y-values. This sketching process further reveals the nature of the quadratic relationship in the function, demonstrating symmetry and its directional opening.