Problem 33 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 29-34: Find the standard equation of a parabola that has a vertical axis and satisfies the given conditions. \(x\)-intercepts \(-3\) and 5 , highest point has \(y\)-coordinate 4

Step-by-Step Solution

Answer

The standard equation is \( y = -\frac{1}{4}x^2 + \frac{1}{2}x + \frac{15}{4} \).

Step 1: Identify the Vertex

Since the parabola has a vertical axis and the highest point is given, this point is the vertex. The vertex form of a parabola is \( y = a(x-h)^2 + k \). Here, \(k=4\) (the highest point's y-coordinate) and \(h\) is the midpoint of the x-intercepts \(-3\) and \(5\), which is \(h = \frac{-3 + 5}{2} = 1\). Thus, the vertex is \((1, 4)\).

Step 2: Use the X-intercepts

The x-intercepts are solutions to the equation \( a(x - (-3))(x - 5) = 0 \) which simplifies to \( a(x+3)(x-5) = 0 \). The expanded form is \( a(x^2 - 2x - 15) = 0 \).

Step 3: Determine the 'a' Value Using the Vertex

Plug in the vertex \((1, 4)\) into the expanded form to find the value of \(a\):\[ 4 = a(1^2 - 2 \cdot 1 - 15) \] \[ 4 = a(1 - 2 - 15) \] \[ 4 = a(-16) \] \[ a = -\frac{1}{4} \].

Step 4: Write the Standard Equation

Substitute \(a = -\frac{1}{4}\) back into the parabola expression. The equation is: \[ y = -\frac{1}{4}(x^2 - 2x - 15) \], which can also be written as \[ y = -\frac{1}{4}x^2 + \frac{1}{2}x + \frac{15}{4} \].

Key Concepts

Vertex Form

The vertex form of a parabola is an alternate way of expressing the quadratic equation. It highlights the vertex, which is the peak or trough of the parabola. Unlike the standard form, which is expressed as \( y = ax^2 + bx + c \), the vertex form is \( y = a(x-h)^2 + k \). Here:

• \( (h, k) \) is the vertex of the parabola.
• \( a \) determines the direction and width of the parabola.

Understanding the vertex form is crucial because it provides easy access to the parabola's vertex, making it simpler to graph and analyze its properties. In our exercise, the parabola's vertex is at \((1, 4)\), which means \( h = 1 \) and \( k = 4 \). This vertex form makes it straightforward to identify that the highest point on the parabola is at \( y = 4 \).

X-intercepts

X-intercepts are the points where the parabola crosses the x-axis. These points are vital in defining the parabola's span across the graph. In the context of a quadratic equation, x-intercepts are found by solving \( ax^2 + bx + c = 0 \). For our exercise, the parabola crosses the x-axis at \(-3\) and \(5\). This gives us the equation \( a(x+3)(x-5) = 0 \) when set to find intercepts.A useful tip is to find the midpoint of these intercepts to locate the x-coordinate of the vertex, which in this case, is \( h = \frac{-3+5}{2} = 1 \). This process directly assists in transforming the standard form into the vertex form by confirming the vertex's x-component.

Equation of a Parabola

The equation of a parabola in standard form is \( y = ax^2 + bx + c \). It offers a straightforward way to plug in values and calculate exact points on the curve. However, converting this to vertex form \( y = a(x-h)^2 + k \) can make understanding the parabola easier.

• The parameter \( a \) is vital as it influences the parabola's orientation and width (a negative \( a \) causes the parabola to open downwards).
• Vertex \((h, k)\) gives information about the highest or lowest point.

In our solution, once we've found \( a = -\frac{1}{4} \) by substituting the vertex back into the factored form, we write the equation as \( y = -\frac{1}{4}(x-1)^2 + 4 \) to highlight the vertex, or alternatively transformed into \( y = -\frac{1}{4}x^2 + \frac{1}{2}x + \frac{15}{4} \). This equation encapsulates the behavior and graphical representation of our parabola.