Problem 39 - Algebra 2

Question

Graph each function. $$ y=\frac{1}{3} x^{2}-4 x+15 $$

Step-by-Step Solution

Answer

The function is a parabola with vertex \((6, 3)\), y-intercept at \((0, 15)\), and opens upward.

Step 1: Identify the function type

The function given is a quadratic function of the form \( y = ax^2 + bx + c \), where \( a = \frac{1}{3} \), \( b = -4 \), and \( c = 15 \). Quadratic functions graph to form a parabola.

Step 2: Determine the vertex

The vertex of a parabola in the form \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Substituting the values: \(-\frac{-4}{2 \times \frac{1}{3}} = 6\). Substitute \( x = 6 \) back into the equation to find \( y \):\[y = \frac{1}{3}(6)^2 - 4(6) + 15 = \frac{1}{3} \times 36 - 24 + 15 = 3 \].Thus, the vertex is \((6, 3)\).

Step 3: Find the y-intercept

To find the y-intercept, set \( x = 0 \) in the equation. This gives:\[ y = \frac{1}{3}(0)^2 - 4(0) + 15 = 15 \].So, the y-intercept is \((0, 15)\).

Step 4: Find the x-intercepts (if any)

To find the x-intercepts, set \( y = 0 \) and solve the equation:\[ 0 = \frac{1}{3}x^2 - 4x + 15 \].Multiply every term by 3 to clear the fraction:\[ x^2 - 12x + 45 = 0 \].Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we have \( a=1, b=-12, c=45 \):\[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(45)}}{2(1)} = \frac{12 \pm \sqrt{144 - 180}}{2} = \frac{12 \pm \sqrt{-36}}{2} \].Since the discriminant is negative, there are no real x-intercepts.

Step 5: Sketch the graph

Plot the vertex \((6, 3)\) on the graph. The y-intercept is at \((0, 15)\). Since the parabola opens upward (as \( a = \frac{1}{3} > 0 \)), it is symmetric about the vertical line through the vertex \( x = 6 \). Since there are no real x-intercepts, the parabola does not cross the x-axis. Draw a smooth curve through the points, emphasizing the upward opening of the parabola.

Key Concepts

Parabola Graphing

Quadratic functions graph into a shape known as a parabola. Parabolas have a unique, symmetric, U-like shape. The direction of this curve depends on the sign of the coefficient 'a' in the quadratic expression. For the function \[ y = \frac{1}{3}x^2 - 4x + 15 \]- Since the coefficient \( a = \frac{1}{3} \) is positive, the parabola opens upwards.- A parabola is symmetric about a vertical line called the axis of symmetry, which passes through the vertex of the parabola.When graphing a quadratic function:- Start by identifying the vertex, the highest or lowest point depending on the graph's orientation.- The y-intercept provides an initial point on the graph.- A parabola's curvature is smooth and continuous, without sharp angles.

Vertex Calculation

The vertex of a parabola is an important feature. It represents either a maximum or minimum point on the graph. To find the vertex from the standard form equation \[ y = ax^2 + bx + c \]you use the formula \[ x = -\frac{b}{2a} \].Substituting for the given function:- The values are \( a = \frac{1}{3} \), \( b = -4 \), and \( c = 15 \).- Calculate \( x \) coordinate: \(-\frac{-4}{2 \times \frac{1}{3}} = 6\).- For the y-coordinate, substitute \( x = 6 \) back into the function: - \( y = \frac{1}{3}(6)^2 - 4(6) + 15 = 3 \).Thus, the vertex of the parabola is \((6, 3)\). This point is vital for determining the graph's symmetry and shape.

Y-Intercept and X-Intercepts

Intercepts are points where the graph crosses the axes. Finding these points helps us understand the position and shape of the parabola.### Y-Intercept- The y-intercept is where the graph crosses the y-axis, so set \( x = 0 \) in the equation: \[ y = \frac{1}{3}(0)^2 - 4(0) + 15 = 15 \].- Thus, the y-intercept is at \((0, 15)\).### X-Intercepts- These are points where the graph crosses the x-axis, found by setting \( y = 0 \) and solving the equation: - Rearrange: \( x^2 - 12x + 45 = 0 \). - The discriminant \( b^2 - 4ac \) is negative, specifically \( \sqrt{-36} \), indicating no real x-intercepts.- Therefore, the parabola does not cross the x-axis. This occurs when a parabola lies entirely above or below the x-axis, confirming no real roots.

Quadratic Formula

The quadratic formula is a crucial tool for finding the roots of any quadratic equation. The formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].- For our equation \( x^2 - 12x + 45 = 0 \), we have \( a = 1 \), \( b = -12 \), and \( c = 45 \). - Plug these values into the formula: \[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(45)}}{2(1)} \]. - Evaluate the discriminant: \( 144 - 180 = -36 \).This negative discriminant means the square root of a negative number, indicating no real solutions for x-intercepts. The knowledge of the quadratic formula thus helps us identify when solutions to a quadratic equation exist in the real number system.