Problem 17 - Algebra 2

Question

Graph each equation. $$ y=x^{2}-12 x+20 $$

Step-by-Step Solution

Answer

Graph the parabola with vertex (6, -16), y-intercept (0, 20), x-intercepts (2, 0) and (10, 0).

Step 1: Identify the type of equation

The given equation is in the form of a quadratic equation, i.e., a polynomial of degree 2 in the form \(y = ax^2 + bx + c\). Here, \(a = 1\), \(b = -12\), and \(c = 20\). Quadratic equations graph as parabolas.

Step 2: Find vertex

The vertex of a quadratic equation \(y = ax^2 + bx + c\) can be found using the formula \(x = -\frac{b}{2a}\). Substitute \(b = -12\) and \(a = 1\):\[x = -\frac{-12}{2 \times 1} = 6\]Plug \(x = 6\) back into the equation to find \(y\):\[y = (6)^2 - 12 \times 6 + 20 = 36 - 72 + 20 = -16\]Thus, the vertex is \((6, -16)\).

Step 3: Find the y-intercept

The y-intercept occurs where \(x = 0\). Substitute \(x = 0\) into the equation:\[y = (0)^2 - 12 \times 0 + 20 = 20\]Thus, the y-intercept is \((0, 20)\).

Step 4: Find x-intercepts

The x-intercepts occur where \(y = 0\). Solve the equation:\[0 = x^2 - 12x + 20\]Use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substituting \(a = 1\), \(b = -12\), and \(c = 20\):\[x = \frac{12 \pm \sqrt{(-12)^2 - 4 \times 1 \times 20}}{2 \times 1} = \frac{12 \pm \sqrt{144 - 80}}{2} = \frac{12 \pm \sqrt{64}}{2}\]\[x = \frac{12 \pm 8}{2}\]This gives the solutions \(x = 10\) and \(x = 2\). Thus, the x-intercepts are \((10, 0)\) and \((2, 0)\).

Step 5: Sketch the graph

1. Plot the vertex \((6, -16)\), the y-intercept \((0, 20)\), and the x-intercepts \((10, 0)\) and \((2, 0)\) on the coordinate plane.2. Draw a smooth curve through these points opening upwards since the coefficient of \(x^2\) is positive, forming a parabola.3. The vertex is a minimum point given the upward opening of the parabola.

Key Concepts

Understanding the Parabola

A parabola is a U-shaped curve that you encounter often in mathematics, especially when dealing with quadratic equations. When you graph a quadratic equation, you always get a parabola. The direction in which this curve opens depends on the coefficient of the squared term in the equation.

If the coefficient of the squared term (\(a\) in \(ax^2 + bx + c\)) is positive, the parabola opens upwards like a regular U. If it is negative, it opens downwards like an upside-down U.

• The highest or lowest point of the parabola is called the vertex.
• Parabolas are symmetrical: they look the same on either side of the vertex.

These properties make parabolas quite predictable in shape, which can be very helpful when solving and graphing quadratic equations.

Finding the Vertex

The vertex of a parabola is its turning point, which is either the highest or the lowest point on the graph. The position of this point gives you valuable information about the function represented by the quadratic equation.

To find the vertex of a parabola given by \(y = ax^2 + bx + c\), you use the vertex formula: \(x = -\frac{b}{2a}\). Substituting this back into the equation gives you the y-coordinate.

• Finding the vertex allows you to know exactly where the parabola changes direction, from falling to rising, or vice versa.
• In the equation \(y = x^2 - 12x + 20\), the vertex is calculated as \((6, -16)\), indicating the lowest point of a parabola that opens upwards.

Exploring the Intercepts

Intercepts are where the parabola crosses the axes. These points can give you important insights into the behavior of the quadratic equation on the graph.

**Y-Intercept:** The y-intercept is found where the parabola crosses the y-axis, which happens when \(x = 0\). You find this point by substituting \(x = 0\) into the equation, giving you the point on the graph where the curve intersects the vertical axis. In our specific equation, the y-intercept is \((0, 20)\).

**X-Intercepts:** X-intercepts occur where the parabola crosses the x-axis, meaning \(y = 0\). Solving the quadratic equation for \(y = 0\) using methods like the quadratic formula reveals these points. They are the roots of the equation. For \(y = x^2 - 12x + 20\), the x-intercepts are at \((10, 0)\) and \((2, 0)\).

• X and y intercepts are crucial for sketching the graph accurately.
• They provide a framework to draw the parabola and understand its extent across the coordinate plane.

Using the Quadratic Formula

The quadratic formula is a powerful tool for finding the roots of any quadratic equation, especially when factoring is difficult. This formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]With this, you can calculate the values of \(x\) where the function equals zero, also known as the x-intercepts.

In the case of \(y = x^2 - 12x + 20\), we used the quadratic formula to find the x-intercepts:\(x = 10\) and \(x = 2\). These values arose from:\(b = -12, a = 1,\) and \(c = 20\).

• The discriminant \(b^2 - 4ac\) inside the formula decides the nature of the roots (real or complex).
• If the discriminant is positive, you get two real roots; if it's zero, one real root; if negative, no real roots.

The quadratic formula is universally applicable and ensures you can solve any quadratic equation systematically.