Problem 16
Question
Compare the graph of the quadratic function with the graph of \(y=x^{2}\). $$f(x)=-\frac{1}{4} x^{2}$$
Step-by-Step Solution
Verified Answer
The function \(f(x)=-\frac{1}{4} x^{2}\) is a downward-opening parabola, wider than the basic function \(y=x^{2}\), which is an upward-opening parabola.
1Step 1: Identify the basic function and its properties
The function \(y=x^{2}\) represents a basic quadratic function. It is a parabola that opens upwards, whose vertex is at the origin (0,0).
2Step 2: Identify the characteristics of the second function
Function \(f(x) = -\frac{1}{4} x^{2}\) is also a quadratic function, but it differs in a few aspects. The minus sign before the fraction indicates that the orientation of the parabola is downward. Also, the coefficient of \(x^{2}\) is \(-\frac{1}{4}\), less than 1, which means the parabola is vertically compressed by a factor of \(\frac{1}{4}\).
3Step 3: Compare the two functions
Both graphs are parabolas, but they open in opposite directions. The graph of the basic function \(y=x^{2}\) opens upwards, while the graph of the function \(f(x)=-\frac{1}{4}x^{2}\) opens downwards due to the negative sign. Furthermore, the function \(f(x)\) has a vertical compression by a factor of \(\frac{1}{4}\), which means it is 'wider' or extends more horizontally compared to the graph of \(y=x^{2}\).
Key Concepts
ParabolaVertical CompressionVertex
Parabola
A parabola is a U-shaped curve that is commonly found in the graphs of quadratic functions. The most basic form of a quadratic function is given by the equation \(y = x^2\). This function forms a parabola that opens upwards, centered around the vertical axis, and has its vertex at the origin (0,0). Parabolas are symmetric and can open upwards or downwards depending on their coefficients.
- If the coefficient of \(x^2\) is positive, the parabola opens upward, like \(y = x^2\).
- If the coefficient of \(x^2\) is negative, the parabola opens downward, as seen in \(f(x) = -\frac{1}{4} x^2\).
Vertical Compression
Vertical compression is a transformation that affects the width of the parabola. It occurs when the absolute value of the coefficient before \(x^2\) in the quadratic function is less than 1. In the function \(f(x) = -\frac{1}{4} x^2\), this coefficient is \(-\frac{1}{4}\). This indicates that the parabola is compressed vertically by a factor of \(\frac{1}{4}\), making it wider than the standard parabola formed by \(y = x^2\).
- The term 'compression' might sound like the graph is being squeezed, but in this case, it means the graph extends more horizontally.
- A vertical compression reduces how steep the parabola appears, effectively making it appear flatter.
Vertex
The vertex of a parabola is the point where it turns; it can be thought of as its highest or lowest point. In the function \(y = x^2\), the vertex is at the origin (0,0). When considering transformations, it is essential to track changes to the vertex since they affect the graph's position.
- A basic transformation that includes a negative sign in front of the quadratic term, such as in \(f(x) = -\frac{1}{4} x^2\), does not affect the vertex position horizontally or vertically.
- In this function, even though the parabola opens downward and is vertically compressed, the vertex remains at (0,0).
Other exercises in this chapter
Problem 16
Use long division to divide. Divisor \(x-2\) Dividend $$x^{3}-x^{2}+2 x-8$$
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Use the graph of \(y=x^{4}\) to sketch the graph of the function. $$f(x)=\frac{1}{2}(x-1)^{4}$$
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Find all the zeros of the function and write the polynomial as a product of linear factors. $$h(x)=x^{3}-3 x^{2}+4 x-2$$
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Write the complex number in standard form and find its complex conjugate. $$(-i)^{3}$$
View solution