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Chapter 2: Problem 17

Find the vertex, the \(x\) -intercepts (if any), and sketch the parabola. \(f(x)=1.2 x^{2}+3.2 x-1.2\)

Short Answer

Expert verified
The vertex of the parabola is \(\left(-\frac{4}{3},-\frac{40}{3}\right)\), the x-intercepts are approximately \((-3.09, 0)\) and \((0.76, 0)\). The parabola opens upward since the coeffcient of the quadratic term is positive. To sketch the parabola, plot the vertex, x-intercepts, and the axis of symmetry, which is at \(x = -\frac{4}{3}\), then draw a U-shaped curve passing through the points.

Step by step solution

01

Rewrite the function in standard form

The given quadratic function is \(f(x) = 1.2x^2 + 3.2x - 1.2\). To rewrite it in standard form, we need to complete the square. To do this, perform the following steps: 1. Factor out the coefficient of the \(x^2\) term from the quadratic and linear terms: \(f(x) = 1.2(x^2 + \frac{8}{3}x) - 1.2\) 2. Divide the linear term coefficient by 2 and square it: \((\frac{8}{3\times2})^2 = \frac{16}{9}\) 3. Add and subtract this value inside the parentheses: \(f(x) = 1.2\left(x^2 + \frac{8}{3}x + \frac{16}{9} - \frac{16}{9}\right) - 1.2\) 4. Write the trinomial as a perfect square: \(f(x) = 1.2\left(\left(x + \frac{4}{3}\right)^2 - \frac{16}{9}\right) - 1.2\) 5. Distribute the 1.2 and simplify the function: \(f(x) = 1.2\left(x + \frac{4}{3}\right)^2 - \frac{40}{3}\) Now the function is in standard form: \(f(x) = 1.2\left(x + \frac{4}{3}\right)^2 - \frac{40}{3}\)
02

Find the vertex of the parabola

The vertex of a quadratic function in standard form \(f(x) = a(x-h)^2 + k\) is given by \((h,k)\). Our function is now written as: \(f(x) = 1.2(x + \frac{4}{3})^2 - \frac{40}{3}\) Therefore, the vertex is at \(\left(-\frac{4}{3},-\frac{40}{3}\right)\).
03

Find the x-intercepts of the parabola

To find the x-intercepts of the parabola, set \(f(x)=0\) and solve for \(x\). \(1.2(x + \frac{4}{3})^2 - \frac{40}{3} = 0\) First, isolate the squared term: \((x + \frac{4}{3})^2 = \frac{40}{3\times1.2}\) Now take the square root of both sides: \(x + \frac{4}{3} = \pm\sqrt{\frac{40}{3\times1.2}}\) Finally, isolate \(x\) and simplify: \(x = -\frac{4}{3} \pm\sqrt{\frac{40}{3\times1.2}}\) Approximately, \(x \approx -3.09, 0.76\). So, the x-intercepts are \((-3.09, 0)\) and \((0.76, 0)\).
04

Sketch the parabola

To sketch the parabola, plot the vertex \(\left(-\frac{4}{3},-\frac{40}{3}\right)\), the x-intercepts \((-3.09, 0)\) and \((0.76, 0)\), and the axis of symmetry, which is a vertical line through the vertex, at \(x = -\frac{4}{3}\). Since the coefficient of the quadratic term is positive, the parabola opens upward. Sketch a U-shaped parabola passing through all the plotted points. Here, we have found the vertex, the x-intercepts, and sketched the parabola for the given quadratic function: \(f(x) = 1.2x^2 + 3.2x - 1.2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertex of a Parabola
The vertex of a parabola is a key point on its curve. It represents the peak if the parabola opens downwards, or the lowest point if it opens upwards. Understanding the vertex helps you graph the parabola effectively.
The standard form of a quadratic function is given as \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex of the parabola. In this exercise, after completing the square, our function becomes \( f(x) = 1.2(x + \frac{4}{3})^2 - \frac{40}{3} \).
Therefore, the vertex is \((-\frac{4}{3}, -\frac{40}{3})\). This point helps in defining the axis of symmetry of the parabola and is crucial when sketching it.
Completing the Square
Completing the square is a method used to transform a quadratic expression into a perfect square trinomial. This is useful for solving quadratic equations and is essential for finding the vertex form of a parabola.
To complete the square:
  • Take the quadratic equation \( ax^2 + bx + c \).
  • Factor out the coefficient \(a\) from the quadratic and linear terms.
  • Next, find \(\left(\frac{b}{2a}\right)^2\), add and subtract this inside the parentheses to form a perfect square.
In our function, we rewrote \( f(x) = 1.2(x^2 + \frac{8}{3}x) - 1.2 \) and found \( \left(\frac{4}{3}\right)^2 \) to complete the square.
X-Intercepts
Finding the x-intercepts involves setting the function equal to zero and solving for \( x \). X-intercepts are the points where the parabola intersects the x-axis. These are crucial for understanding the parabola's position relative to the axes.
For the function \( f(x) = 1.2(x + \frac{4}{3})^2 - \frac{40}{3} \), set \( f(x) = 0 \). After isolating and simplifying:
  • \( (x + \frac{4}{3})^2 = \frac{40}{3 \times 1.2} \)
  • Solve for \( x \): \( x = -\frac{4}{3} \pm \sqrt{\frac{40}{3 \times 1.2}} \)
The solutions approximately are \( x \approx -3.09 \) and \( x \approx 0.76 \). Hence, the x-intercepts are \((-3.09, 0)\) and \((0.76, 0)\).
Graphing Parabolas
Graphing a parabola involves plotting certain key points on a coordinate grid to paint a picture of its shape. Important aspects include the vertex, x-intercepts, and the direction in which it opens.
For this function, plot the vertex \((-\frac{4}{3}, -\frac{40}{3})\) and the x-intercepts \((-3.09, 0)\) and \((0.76, 0)\). The axis of symmetry for the parabola is a vertical line through the vertex, at \( x = -\frac{4}{3} \), dividing the parabola into mirror images.
Because the coefficient of the quadratic term \(1.2\) is positive, the parabola opens upward. You'll end up with a U-shaped graph, providing a clear visual representation of the quadratic function.

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