Problem 19
Question
What is the effect on the graph of the equation \(y=x^{2}+2\) when the equation is changed to \(y=x^{2}-5 ?\)
Step-by-Step Solution
Verified Answer
The graph shifts 7 units down.
1Step 1: Identify Original Graph
The original equation is \(y = x^2 + 2\). This represents a parabola that opens upwards with vertex at the point \((0, 2)\). This is the standard form of a quadratic equation \(y = x^2\) shifted 2 units up along the y-axis.
2Step 2: Identify New Graph
The new equation is \(y = x^2 - 5\). This is also a parabola that opens upwards, but its vertex is at the point \((0, -5)\). This equation represents a vertical shift of the original graph 7 units down.
3Step 3: Determine the Effect of the Change
By comparing the two equations \(y = x^2 + 2\) and \(y = x^2 - 5\), we notice that the change from \(+2\) to \(-5\) in the constant term results in shifting the original graph downward by 7 units along the y-axis.
Key Concepts
Quadratic FunctionsVertex FormVertical Shift
Quadratic Functions
Quadratic functions are a fundamental concept in algebra and pre-calculus. They are defined by equations of the form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. The graph of a quadratic function is called a parabola.
Parabolas are unique in that they have a symmetrical shape and can open upwards or downwards. If the coefficient \(a\) is positive, the parabola opens upwards, resembling a U-shape. If \(a\) is negative, it opens downwards. For any parabola, a key feature is its vertex, which represents the highest or lowest point on the graph, depending on the direction it opens.
Understanding the structure and transformation of quadratic functions is crucial as they not only apply to pure math problems but also to real-world phenomena such as projectile motion and optimizing areas and volumes.
Parabolas are unique in that they have a symmetrical shape and can open upwards or downwards. If the coefficient \(a\) is positive, the parabola opens upwards, resembling a U-shape. If \(a\) is negative, it opens downwards. For any parabola, a key feature is its vertex, which represents the highest or lowest point on the graph, depending on the direction it opens.
Understanding the structure and transformation of quadratic functions is crucial as they not only apply to pure math problems but also to real-world phenomena such as projectile motion and optimizing areas and volumes.
Vertex Form
The vertex form of a quadratic function offers a convenient way to identify the vertex of the parabola. This form is expressed as \(y = a(x-h)^2 + k\), where \((h, k)\) is the vertex point of the parabola.
By using the vertex form, we can easily see how the graph translates on the coordinate plane. The value of \(h\) indicates a horizontal shift along the x-axis, while \(k\) shows a vertical shift along the y-axis.
By using the vertex form, we can easily see how the graph translates on the coordinate plane. The value of \(h\) indicates a horizontal shift along the x-axis, while \(k\) shows a vertical shift along the y-axis.
- For example, an equation \(y = (x-3)^2 + 4\) corresponds to a parabola with its vertex at \((3, 4)\).
- Changes in \(a\) affect the width and direction of the parabola's opening.
Vertical Shift
Vertical shifts occur when we add or subtract a constant to or from the quadratic function. This transformation affects the graph by moving it up or down the y-axis without altering its shape.
A positive constant added moves the graph upwards, while a negative constant shifts it downwards. For instance, if you begin with the quadratic equation \(y = x^2 + 2\), adding 2 results in lifting the entire parabola two units upward, positioning the vertex at \((0, 2)\).
Alternatively, changing the equation to \(y = x^2 - 5\) indicates a vertical shift 7 units downward—this is because the graph moves from 2 units above the x-axis to 5 units below.
A positive constant added moves the graph upwards, while a negative constant shifts it downwards. For instance, if you begin with the quadratic equation \(y = x^2 + 2\), adding 2 results in lifting the entire parabola two units upward, positioning the vertex at \((0, 2)\).
Alternatively, changing the equation to \(y = x^2 - 5\) indicates a vertical shift 7 units downward—this is because the graph moves from 2 units above the x-axis to 5 units below.
- Vertical shifts do not affect the x-coordinate of the vertex, only the y-coordinate.
- Such shifts maintain the 'U' shape of parabolas but reposition them vertically.
Other exercises in this chapter
Problem 19
Complete parts a–c for each quadratic equation. a. Find the value of the discriminant. b. Describe the number and type of roots. c. Find the exact solutions by
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Simplify. $$ (1+2 i)(-1+4 i) $$
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Solve each equation by using the Square Root Property. \(x^{2}+7 x+\frac{49}{4}=4\)
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