Problem 4 - Algebra 2

Question

Graph each inequality. $$ y \leq-x^{2}+5 x+6 $$

Step-by-Step Solution

Answer

Graph the downward-opening parabola with vertex \( \left(\frac{5}{2}, \frac{17}{4}\right) \), shade below it for \( y \leq -x^2 + 5x + 6 \).

Step 1: Identify the Type of Conic Section

The given inequality is quadratic in form because it involves a term with \( x^2 \). Quadratic equations represent parabolas. The general form of a quadratic equation is \( y = ax^2 + bx + c \). Here, \( a = -1 \), \( b = 5 \), and \( c = 6 \). Since \( a < 0 \), the parabola opens downwards.

Step 2: Find the Vertex of the Parabola

The vertex of a parabola given by \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Substituting \( a = -1 \) and \( b = 5 \), we get \( x = -\frac{5}{2(-1)} = \frac{5}{2} \). Substitute \( x = \frac{5}{2} \) back into the equation to find \( y \), the vertex coordinate: \( y = -\left(\frac{5}{2}\right)^2 + 5\left(\frac{5}{2}\right) + 6 \). After simplifying, \( y = \frac{17}{4} \). Therefore, the vertex is \( \left(\frac{5}{2}, \frac{17}{4}\right) \).

Step 3: Determine Intercepts

The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the equation: \( y = -(0)^2 + 5(0) + 6 = 6 \). The y-intercept is \( (0, 6) \). To find the x-intercepts, solve \( 0 = -x^2 + 5x + 6 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For this equation, \( a = -1 \), \( b = 5 \), \( c = 6 \). Solving gives the x-intercepts as \( x = -1 \) and \( x = 6 \).

Step 4: Graph the Parabola

Using the vertex \( \left(\frac{5}{2}, \frac{17}{4}\right) \), y-intercept \( (0, 6) \), and x-intercepts \( (-1, 0) \) and \( (6, 0) \), sketch the parabola. The parabola opens downwards.

Step 5: Shade the Region

For the inequality \( y \leq -x^2 + 5x + 6 \), you need to shade the region below the parabola because it indicates where the y-values are less than or equal to the quadratic expression. Solid lines are used because the inequality is \( \leq \) (inclusive). The resulting graph represents all points \( (x, y) \) for which \( y \leq -x^2 + 5x + 6 \).

Key Concepts

Graphing Parabolas

When dealing with quadratic inequalities, understanding the graph of a related quadratic equation is crucial. A quadratic equation forms a parabola. The basic parental form is \( y = ax^2 + bx + c \). The sign of \( a \) determines the direction it opens:

• If \( a > 0 \), the parabola opens upwards.
• If \( a < 0 \), the parabola opens downwards, as in our case where \( a = -1 \).

To graph the parabola, plot key points like the vertex, x-intercepts, and the y-intercept, then draw a smooth curve through these points. Ensure the curve is symmetrical about the vertex.

Vertex of a Parabola

The vertex of a parabola is an important point that provides information on its maximum or minimum value. It lies at the axis of symmetry of the parabola.
To find the vertex, use the formula for the x-coordinate: \( x = -\frac{b}{2a} \). Substituting our values, we have \( x = -\frac{5}{2(-1)} = \frac{5}{2} \).Plug this x-value back into the quadratic equation to find the corresponding y-coordinate, yielding \( y = \frac{17}{4} \).
Thus, the vertex in this exercise is \( \left(\frac{5}{2}, \frac{17}{4}\right) \). This vertex is a maximum point for the downward-opening parabola.

Intercepts of Parabolas

Intercepts are the points where the parabola crosses the axes.

• The **y-intercept** is where the graph crosses the y-axis. For this point, set \( x = 0 \) and solve for \( y \). Our calculation gives only one y-intercept: \( (0, 6) \).
• The **x-intercepts** occur where the graph crosses the x-axis. Solve the equation \( 0 = -x^2 + 5x + 6 \) using the quadratic formula. The solutions give us the x-intercepts \( x = -1 \) and \( x = 6 \).

These intercepts are useful for drawing the parabola as they provide more points through which the parabola must pass.

Shading Regions on Graphs

When graphing inequalities like \( y \leq -x^2 + 5x + 6 \), shading indicates the solution region.To determine which side of the parabola to shade, use a test point. Usually, the parabola itself helps:

• If the inequality is \( \leq \) or \( \geq \), shade *below* (for \( \leq \)) or *above* (for \( \geq \)), and draw the parabola with a solid line.
• If the inequality is \( < \) or \( > \), shade outside the parabola boundaries, and use a dashed line.

For our inequality, the shading is below the parabola, with solid lines indicating inclusive boundaries. This shaded region includes all points \( (x, y) \) where the y-value meets the inequality condition and reflects the solution set of the inequality.