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Chapter 2: Problem 13

Find the vertex, the \(x\) -intercepts (if any), and sketch the parabola. \(f(x)=16-x^{2}\)

Short Answer

Expert verified
Vertex: \((0,16)\) X-Intercepts: \((-4,0)\) and \((4,0)\) Sketch: The parabola opens downwards, has its vertex at \((0,16)\), and passes through the x-intercepts \((-4,0)\) and \((4,0)\).

Step by step solution

01

1. Vertex

To find the vertex of the function, we need to put the quadratic function in vertex form, which looks like this: \[f(x) = a(x-h)^2 + k\] where (h,k) is the vertex of the parabola. The given function is: \[f(x) = 16-x^2\] This is already in vertex form, with \(a = -1\), \(h = 0\), and \(k = 16\). Thus, the vertex of the parabola is \((0,16)\).
02

2. X-Intercepts

To find the x-intercepts, we set f(x) equal to zero and solve for x: \[0 = 16 - x^2\] Now we need to solve for x: \[x^2 = 16\] To solve for x, take the square root of both sides: \[x = \pm\sqrt{16}\] This gives us the two x-intercepts when x is positive and when x is negative: \[x = -4, 4\] So, the x-intercepts are \((-4,0)\) and \((4,0)\).
03

3. Sketch the Parabola

The parabola will open downwards since the coefficient of the x-squared term is negative. Let's plot the vertex, which is the highest point on the parabola, and the x-intercepts: - Vertex: \((0,16)\) - X-intercepts: \((-4,0)\) and \((4,0)\) Now, we can sketch the parabola using these points. Place the vertex at the highest point and make sure the parabola passes through both x-intercepts. The arms of the parabola will be symmetrical and opening downwards.

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Key Concepts

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

Vertex Form
In quadratic functions, the vertex form is a widely used representation. The vertex form of a quadratic function is given by:\[ f(x) = a(x-h)^2 + k \]where:
  • \(a\) is a constant that determines the direction and the "width" of the parabola.
  • \((h, k)\) is the vertex of the parabola.
This form is helpful because it directly provides the coordinates of the vertex, making it easier to analyze the parabola's most extreme point.
In our example, the function is already rewritten as:\[ f(x) = 16 - x^2 \]We can easily rewrite this (noticing it's equivalent to the form) as:\( f(x) = -(x-0)^2 + 16 \) This tells us that the vertex is located at \( (0, 16) \). The coefficient \( a = -1 \) indicates that the parabola opens downwards.
Parabola Sketching
Sketching a parabola involves plotting its shape based on certain key features. The vertex is often the starting point for sketching.
Here鈥檚 how to sketch the parabola for our specific function:
  • Identify the vertex, which is \( (0, 16) \). This point is the peak of the parabola since the value of \( a \) is negative, meaning it opens downwards.
  • Locate the x-intercepts, \( (-4, 0) \) and \( (4, 0) \), which are the points where the parabola crosses the x-axis.
  • Understand that the parabola is symmetric with respect to the vertical line passing through the vertex, known as the axis of symmetry.
The parabola opens downwards due to the negative \( a \), making the vertex the parabola's highest point.
Connecting these plotted points will give a u-shaped curve opening downwards, completing the sketch.
X-Intercepts
The x-intercepts are the points where the graph of the quadratic function crosses the x-axis. This occurs when \( f(x) = 0 \).
To find the x-intercepts, set the function equal to zero and solve for \( x \):\[ 0 = 16 - x^2 \]Solving this, we isolate \( x^2 \):\[ x^2 = 16 \]Taking the square root of both sides gives:\[ x = \pm \sqrt{16} \]Thus, \( x = -4 \) and \( x = 4 \). These are the x-intercepts \( (-4,0) \) and \( (4,0) \).
The x-intercepts are essential for sketching the parabola, as they confirm where the curve touches or crosses the x-axis, forming a clear anchor for the shape and direction of the curve.

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