Problem 22 - College Algebra and Calculus: An Applied Approach

Question

Sketch the graph of the quadratic function. Identify the vertex and intercepts. $$h(x)=x^{2}-9$$

Step-by-Step Solution

Answer

The vertex of the parabola \(h(x)=x^{2}-9\) is at (0, -9). It opens upward, and its x-intercepts are (-3, 0) and (3,0). The y-intercept is at (0, -9), which is also the vertex of the parabola.

Step 1: Identify the Vertex

The vertex of a parabola \(y = ax^{2} + bx + c\) is given by the point \(-b/2a, f(-b/2a)\). For the given function \(h(x) = x^{2}-9\), it is in the form \(y = x^{2} + 0x - 9\), hence having a = 1, b = 0, and c = -9. Therefore, the vertex here is at (0, -b/2a) which equals to (0, -0/2*1) = (0, 0). Substituting x=0 in the given equation gives us the y-coordinate as -9. Hence, the vertex of the parabola is (0, -9)

Step 2: Identify the Orientation & Shape

The given function, \(h(x)=x^{2}-9\), has a positive coefficient for x^2. Thus, the graph of this equation will be a parabola that opens upward. This is because, for a general equation \(ax^{2}+bx+c\), if a > 0 the parabola opens upward and if a < 0, the parabola opens downward. Thus, here the vertex (0, -9) is the lowest point on the graph.

Step 3: Identify the Intercepts

To find x-intercepts, we set \(h(x) = x^{2}-9 = 0\). Solving this equation gives x-intercepts as x = -3, 3 (by factoring). For y-intercept, we set x = 0, thus from the equation \(h(x) = x^{2}-9\),we get y = -9, and hence, the y-intercept is (0, -9)

Step 4: Sketch the Graph

Place the vertex on the graph at point (0, -9). The parabola will open upward, and x-intercepts are (-3, 0) and (3,0). The y-intercept is at (0, -9), which is also the vertex of the parabola. Thus, the graph is a U-shape curve opening upwards with lowest point (vertex) at (0, -9) and crossing x-axis at x=-3 and x=3

Key Concepts

Vertex of a Parabola

The vertex is the pivotal point of a parabola. It serves as either its highest or lowest point, depending on the parabola's orientation. For a standard quadratic equation such as \( y = ax^2 + bx + c \), the vertex can be found using the coordinates \( \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right) \). In our function \( h(x) = x^2 - 9 \):

• It can be rewritten as \( h(x) = x^2 + 0x - 9 \), meaning \( a = 1 \), \( b = 0 \), and \( c = -9 \).
• The x-coordinate of the vertex is \( x = -\frac{0}{2 \times 1} = 0 \).
• Plugging x back into the original function tells us \( y = 0^2 - 9 = -9 \), so the vertex is (0, -9).

This vertex represents the minimum point on the graph, as the parabola will open upwards.

Parabola Orientation

In the context of quadratic functions, the coefficient of the \( x^2 \) term tells us about the parabola's orientation:

• If \( a > 0 \), the parabola opens upwards like a 'U'.
• If \( a < 0 \), it opens downwards like an 'n'.

For our function \( h(x) = x^2 - 9 \), \( a = 1 \), a positive number, determining that the parabola opens upward. Visualize this as a bowl shape. The vertex (0, -9) is the lowest point on this graph, ensuring that all other points on the graph are equal to or above this point.

Intercepts of a Graph

Intercepts are where the graph crosses the axes, revealing critical information:

• The x-intercepts occur where the graph intersects the x-axis. Set the equation to zero, \( x^2 - 9 = 0 \), and solve for \( x \). This results in \( x=3 \) or \( x=-3 \), so the points are (-3, 0) and (3, 0).
• The y-intercept is where the graph intersects the y-axis. Set \( x=0 \) in the function to find \( y \). So, \( h(x) = 0^2 - 9 = -9 \). The y-intercept is at (0, -9), which also turns out to be the vertex in this case.

Both intercepts are essential for sketching an accurate graph, as they act as points of reference for plotting the parabola on a coordinate grid.