Problem 29 - Elementary Algebra

Question

Graph. Find the vertex and the y-intercept. In addition, find the \(x\) - intercepts if they exist. $$ y=x 2-9 $$

Step-by-Step Solution

Answer

Vertex: (0, -9); Y-intercept: (0, -9); X-intercepts: (3, 0) and (-3, 0).

Step 1: Rewrite the Equation

First, make sure the given equation is in the standard form of a quadratic equation. The given equation is \( y = x^2 - 9 \). This is already in a suitable form \( y = ax^2 + bx + c \) with \( a = 1 \), \( b = 0 \), and \( c = -9 \).

Step 2: Find the Vertex

The vertex form of a quadratic equation is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. For a quadratic equation in the form of \( y = ax^2 + bx + c \), the x-coordinate of the vertex \( h \) is found using \( h = -\frac{b}{2a} \). Here, \( b = 0 \) and \( a = 1 \), so \( h = -\frac{0}{2 \cdot 1} = 0 \). Substitute \( x = 0 \) back into the equation to find \( k \). \( y = 0^2 - 9 = -9 \). So, the vertex \( (h, k) = (0, -9) \).

Step 3: Find the Y-intercept

The y-intercept is found by setting \( x = 0 \) in the equation \( y = x^2 - 9 \). Plug \( x = 0 \) into the equation: \( y = 0^2 - 9 = -9 \). Thus, the y-intercept is \( (0, -9) \).

Step 4: Find the X-intercepts

The x-intercepts are found by setting \( y = 0 \) in the equation \( y = x^2 - 9 \). Set \( x^2 - 9 = 0 \). Solve for \( x^2 = 9 \). Taking the square root on both sides, \( x = \pm 3 \). Therefore, the x-intercepts are \( (3, 0) \) and \( (-3, 0) \).

Key Concepts

Vertex

In the context of graphing quadratics, identifying the vertex is a crucial step. The vertex is the point where the parabola, which is the graph of a quadratic equation like \( y = x^2 - 9 \), changes direction. In this equation, the standard form is \( y = ax^2 + bx + c \), where \( a = 1 \), \( b = 0 \), and \( c = -9 \). To find the x-coordinate of the vertex \( h \), use the formula \( h = -\frac{b}{2a} \). Since \( b = 0 \), substituting gives \( h = -\frac{0}{2 \cdot 1} = 0 \).
To find the y-coordinate \( k \), substitute \( h = 0 \) back into the equation: \( y = 0^2 - 9 = -9 \).
Thus, the vertex is the point \( (0, -9) \). You'll notice that this is also the lowest point on the graph when \( a > 0 \), indicating the parabola opens upwards.

Y-intercept

The y-intercept of a quadratic graph is the point where the graph crosses the y-axis. To find this point, set \( x = 0 \) in the original equation \( y = x^2 - 9 \).
Performing the substitution: \( y = 0^2 - 9 = -9 \).
Therefore, the y-intercept is \( (0, -9) \). Since the y-intercept involves setting \( x = 0 \), it is straightforward to compute and reveals crucial information about where the parabola is aligned relative to the y-axis. This point is particularly vital as it usually sets the initial point for graphing the quadratic equation.

X-intercepts

X-intercepts represent the points where the graph intersects the x-axis. These are locations where the value of \( y \) is zero. To find the x-intercepts in the equation \( y = x^2 - 9 \), set the equation to zero: \( x^2 - 9 = 0 \).
Solve for \( x \) by isolating \( x^2 \): \( x^2 = 9 \).
Taking the square root of both sides gives \( x = \pm 3 \).
Thus, there are two x-intercepts: \( (3, 0) \) and \( (-3, 0) \).
These points mark where the quadratic graph crosses the x-axis, determining the roots of the equation. Identifying them helps understand the spread and symmetry of the parabola.