Problem 8 - Calculus Early Transcendentals

Question

Find the vertex, focus, and directrix of the parabola and sketch its graph. $$y+12 x-2 x^{2}=16$$

Step-by-Step Solution

Answer

The vertex is (3, 34), the focus is (3, 271/8), and the directrix is y = 273/8.

Step 1: Rearrange the Equation

Start by rearranging the given equation \[ y + 12x - 2x^2 = 16 \]so that the quadratic term is isolated. Rewrite it in this form:\[ y = 2x^2 - 12x + 16 \].

Step 2: Complete the Square

Next, complete the square for the quadratic expression in \(x\). Begin by factoring out the coefficient of \(x^2\):\[ y = -2(x^2 - 6x) + 16 \].To complete the square, take half of the \(x\)-coefficient, \(-6\), which is \(-3\), square it to get \(9\), and adjust the equation:\[ y = -2(x^2 - 6x + 9 - 9) + 16\]. Factor the perfect square trinomial:\[ y = -2((x-3)^2 - 9) + 16 \].Simplify the expression:\[ y = -2(x-3)^2 + 18 + 16 \].Finally, write it as:\[ y = -2(x-3)^2 + 34 \].

Step 3: Identify the Vertex Form

The equation \( y = -2(x-3)^2 + 34 \) is in the vertex form \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex. Hence, the vertex of the parabola is \((3, 34)\).

Step 4: Determine the Direction of the Parabola

Since the coefficient \(a = -2\) is negative, the parabola opens downwards.

Step 5: Calculate the Focus of the Parabola

The focus of a parabola \( y = a(x-h)^2 + k \) is determined by the formula \( (h, k + \frac{1}{4a}) \). For \(a = -2\), find \( \frac{1}{4(-2)} = -\frac{1}{8} \). Thus, the focus is located at \( (3, 34 - \frac{1}{8}) = (3, \frac{271}{8}) \).

Step 6: Find the Directrix

The directrix of the parabola is given by \( y = k - \frac{1}{4a} \). Using the calculated \( a \), the directrix is:\[ y = 34 + \frac{1}{8} = \frac{273}{8} \].

Step 7: Sketch the Graph

To sketch the parabola, plot its vertex \((3, 34)\), its focus \((3, \frac{271}{8})\), and its directrix \(y = \frac{273}{8}\). The parabola opens downward, curving away from its vertex along the line parallel to the directrix.

Key Concepts

Vertex

The vertex of a parabola is a key point where the direction changes. It is the point that is either the highest or lowest on the curve, depending on whether the parabola opens upwards or downwards. In the vertex form of a quadratic equation, \[ y = a(x-h)^2 + k \], the vertex is given by the point \((h, k)\). Here, \

• \(h\) is the x-coordinate of the vertex,
• \(k\) is the y-coordinate of the vertex.

Once you've completed the square for the equation \( y = -2(x-3)^2 + 34 \), we can directly see that the vertex is located at \((3, 34)\). The vertex not only indicates the peak or lowest point of the parabola but also splits the parabola into two symmetrical halves. When plotting, it's essential to identify the vertex first, as it helps frame the overall shape of the graph.

Focus

The focus of a parabola is a special point located inside the curve. All points on the parabola are equidistant from the focus and a line called the directrix.To find the focus, use the formula:\[ (h, k + \frac{1}{4a}) \]for a parabola given by \( y = a(x-h)^2 + k \). Here,

• \((h, k)\) is the vertex.
• \(a\) is the coefficient that determines the width and direction of the parabola.

In our exercise, \( a = -2 \), so we calculate the shift for the y-coordinate:\[ \frac{1}{4(-2)} = -\frac{1}{8} \]. This results in the focus being at the point:\( (3, 34 - \frac{1}{8}) = (3, \frac{271}{8}) \).The focus helps in understanding how the parabola directs its curves and towards which point it tightens. It's the spotlight of sorts for this curve and aids in sketching and visualizing the parabola's true shape.

Directrix

The directrix is a horizontal line situated opposite the focus, aligned to help balance the parabolic curve.For a parabola in the format \( y = a(x-h)^2 + k \), the directrix is found using:\[ y = k - \frac{1}{4a} \]. Since the parabola opens downward, the directrix will be above the vertex. Using the given vertex \((3, 34)\) and \( a = -2 \), the calculation for the directrix results in:\[ y = 34 + \frac{1}{8} = \frac{273}{8} \].

• This means every point on the parabola is equidistant from the focus and this computed y-value of the directrix.

Understanding the position of the directrix is instrumental when it comes to plotting and graph interpretation, giving a clear guide to the line that the parabola diverges from.

Completing the Square

Completing the square is an algebraic technique used to rearrange a quadratic equation into vertex form. This makes it easier to identify the vertex of a parabola.In our parabola scenario \( y = -2x^2 + 12x - 16 \), we began by rearranging to isolate the quadratic term: \( y = -2(x^2 - 6x) + 16 \).

• Half of the coefficient of \(x\) is \(-3\). Squaring it gives us 9.
• Add and subtract 9 inside the bracket: \( y = -2(x^2 - 6x + 9 - 9) + 16 \).
• This becomes \( y = -2((x-3)^2 - 9) + 16 \).
• Simplifying, this results in \( y = -2(x-3)^2 + 34 \).

This reformulation allows us to see the vertex at \((3, 34)\). Completing the square offers precision in transforming raw equations into a format revealing critical features of the parabola. Understanding this process is crucial for proper graph plotting and interpretation.