Problem 24 - Algebra 2

Question

For Exercises \(24-27,\) use the equation \(x=3 y^{2}+4 y+1\) Draw the graph. Find the \(x\) -intercept \((\mathrm{s})\) and \(y\) -intercept \((\mathrm{s})\)

Step-by-Step Solution

Answer

X-intercept is (1, 0); Y-intercepts are (0, -1) and (0, -1/3).

Step 1: Understand the Equation

The given equation is a quadratic in terms of the variable \( y \): \( x = 3y^2 + 4y + 1 \). We need to graph it and find its intercepts.

Step 2: Finding the x-intercepts

The x-intercepts occur when \( y = 0 \). Substitute \( y = 0 \) into the equation to find \( x \): \[ x = 3(0)^2 + 4(0) + 1 = 1 \]. So the x-intercept is \((1, 0)\).

Step 3: Finding the y-intercepts

The y-intercept occurs where the curve intersects the vertical axis \( y \). For this equation, we substitute \( x = 0 \) and find \( y \): \[ 0 = 3y^2 + 4y + 1 \]. Solving \( 3y^2 + 4y + 1 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 3 \), \( b = 4 \), and \( c = 1 \), gives \[ y = \frac{-4 \pm \sqrt{16 - 12}}{6} = \frac{-4 \pm 2}{6} \]. Thus, \( y = -1 \) or \( y = -\frac{1}{3} \). This results in y-intercepts at \((0, -1)\) and \((0, -\frac{1}{3})\).

Step 4: Sketching the Graph

Since \( x \) is expressed in terms of \( y \), this equation represents a parabola opening sideways (to the right if positive). Plotting the x-intercept \((1, 0)\) and y-intercepts \((0, -1)\) and \((0, -\frac{1}{3})\), we sketch the parabola, ensuring it passes through those points. The vertex is halfway between the y-intercepts; calculate the midpoint \( y = -\frac{1 + (-\frac{1}{3})}{2} = -\frac{2}{3} \) and substitute back into the equation for precise graph drawing.

Key Concepts

Intercepts

Intercepts are the points where a graph crosses the axes. For the given quadratic equation, we identify both the x-intercept and y-intercept. These are crucial as they help us understand the position and direction of the parabola.

**x-intercept**: This is where the graph crosses the x-axis. Since the x-axis is where the y-value is zero, we calculate it by setting \( y = 0 \) and solving for \( x \). For our equation \( x = 3y^2 + 4y + 1 \), the x-intercept is calculated as follows:

• Substitute \( y = 0 \) into the equation: \( x = 3 \cdot 0^2 + 4 \cdot 0 + 1 = 1 \).
• Thus, the x-intercept is at \((1, 0)\).

**y-intercept**: Conversely, this is where the graph crosses the y-axis. At this point, \( x = 0 \). To find the y-intercept, set \( x = 0 \) and solve for \( y \). This requires solving the quadratic equation \( 3y^2 + 4y + 1 = 0 \).

• Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), substituting \( a = 3 \), \( b = 4 \), \( c = 1 \), results in solutions \( y = -1 \) and \( y = -\frac{1}{3} \).
• These result in y-intercepts at \((0, -1)\) and \((0, -\frac{1}{3})\).

Intercepts provide quick visual cues in graphing and interpreting a parabola's behavior.

Graphing Parabolas

Graphing a parabolic curve involves several steps that help us accurately visualize the equation on a coordinate plane.

Since the equation \( x = 3y^2 + 4y + 1 \) is quadratic in terms of \( y \), it represents a parabola. Given that \( x \) is the dependent variable, the parabola opens sideways.

**Key steps for plotting the parabolic curve:**

• Identify the intercepts as starting points. Plot \((1, 0)\), \((0, -1)\), and \((0, -\frac{1}{3})\) on the graph.
• Locate the vertex. The vertex of the parabola is midway between the y-intercepts. Calculate the midpoint: \( y = -\frac{2}{3} \), then substitute back into the equation to find corresponding \( x \).
• Sketch the parabola. Since \( x = 3y^2 + 4y + 1 \), it opens to the right, as \( x \) increases as \( y \) changes.

The curve smoothly passes through all intercepts and arches in the calculated orientation. Graphing reinforces our understanding of the parabola's symmetry and range in the coordinate plane.

Quadratic Formula

The quadratic formula is a widely used tool in solving quadratic equations like \( ax^2 + bx + c = 0 \). It's particularly useful when factoring is cumbersome or impossible.

For our exercise, the quadratic equation to find y-intercepts is \( 3y^2 + 4y + 1 = 0 \). We applied the quadratic formula:

• **Formula**: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Substitute the coefficients: \( a = 3, \; b = 4, \; c = 1 \).

Calculate the discriminant \( b^2 - 4ac \). This discriminant determines the nature and number of solutions:

• If positive, two real distinct solutions exist. Here: \( 16 - 12 = 4 \), resulting in two solutions.
• If zero, exactly one real solution due to multiplicity.
• If negative, no real solutions are possible, indicating imaginary roots.

Substituting into the formula gives solutions \( y = -1 \) and \( y = -\frac{1}{3} \). Through this approach, the quadratic formula helps find precise y-intercepts, providing a complete picture of the parabola’s behavior.