Problem 79 - Intermediate Algebra

Question

Sketch the graph of each equation. If the graph is a parabola, find its vertex. If the graph is a circle, find its center and radius. $$y=5 x^{2}-20 x+16$$

Step-by-Step Solution

Answer

The graph is a parabola with vertex at (2, -4).

Step 1: Identify the Graph Type

The given equation is \( y = 5x^2 - 20x + 16 \), which is a quadratic equation in the form of \( y = ax^2 + bx + c \). This type of equation represents a parabola.

Step 2: Find the Vertex Formula

The vertex of a parabola given by \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex: \( x = -\frac{b}{2a} \).

Step 3: Calculate the x-coordinate of the Vertex

For the equation \( y = 5x^2 - 20x + 16 \), we have \( a = 5 \) and \( b = -20 \). Substitute these values into the vertex formula: \( x = -\frac{-20}{2 \times 5} = 2 \).

Step 4: Calculate the y-coordinate of the Vertex

Substitute \( x = 2 \) back into the original equation to find the y-coordinate: \( y = 5(2)^2 - 20(2) + 16 \). Simplify to get \( y = 20 - 40 + 16 = -4 \).

Step 5: Write the Vertex Coordinates

Combine the x-coordinate and y-coordinate to state the vertex of the parabola. The vertex is \( (2, -4) \).

Step 6: Sketch the Parabola

To sketch the parabola, plot the vertex \( (2, -4) \). The parabola opens upwards because the coefficient \( a = 5 \) is positive. Draw a symmetric curve around the vertex.

Key Concepts

Vertex of a Parabola

In the world of parabolas, the vertex stands as the most crucial point. It's where the parabola changes direction.
To find the vertex of a parabola given by the quadratic equation \( y = ax^2 + bx + c \), you use a simple formula for the x-coordinate: \( x = -\frac{b}{2a} \). This formula is derived from the process of completing the square, a method used to turn a quadratic equation into its vertex form.
Here's how it works:

• First, identify the coefficients \( a \) and \( b \) from the equation. For example, in \( y = 5x^2 - 20x + 16 \), \( a = 5 \) and \( b = -20 \).
• Substitute these values into the formula: \( x = -\frac{-20}{2 \times 5} = 2 \).

Once you have the x-coordinate, you find the y-coordinate by substituting the x back into the original equation. Completing these steps gives you the coordinates of the vertex, \( (x, y) \), which represents the peak or the lowest point of the parabola.

Quadratic Equation

Quadratic equations form the basis of parabolas and many real-world phenomena. They are equations of the second degree, meaning they have a variable squared, usually in the form \( y = ax^2 + bx + c \). Here, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). This form is known as the standard form of a quadratic equation.
The characteristics of a quadratic equation include:

• The direction of the parabola, determined by the sign of \( a \). If \( a > 0 \), the parabola opens upwards (like a smile); if \( a < 0 \), it opens downwards (like a frown).
• The vertex, as discussed previously, which can be seen as the turning point of the parabola.
• The axis of symmetry, a vertical line that passes through the vertex, ensuring the parabola is symmetrical on either side.

These features help us understand and graph quadratic functions, making them easier to analyze.

Graph of a Quadratic Function

Graphing a quadratic function involves plotting its parabolic shape. To do this, you need some essential components determined from its quadratic equation:
The following steps are key in sketching the graph:

• Identify the Vertex: Use the vertex formula to find this point, as it dictates the parabola's position and shape.
• Plot Additional Points: Choose x-values on either side of the vertex to establish symmetry and shape.
• Draw the Parabola: Use the vertex and the plotted points to draw a smooth, U-shaped curve. Ensure it opens in the direction indicated by the sign of \( a \).
• Locate the Y-intercept: This point is where the parabola crosses the y-axis, simply \( c \) from the equation when \( x = 0 \).

A comprehensive graph not only shows the parabola's characteristic shape but also highlights how it can represent various real-world scenarios, from projectile motion to optimization problems.