Chapter 1: Problem 56
Graph the parabolas in Exercises 53–60. Label the vertex, axis, and intercepts in each case. $$ y=-x^{2}+4 x-5 $$
Short Answer
Expert verified
The vertex is \((2, -1)\), axis of symmetry \(x = 2\), y-intercept \((0, -5)\), no real x-intercepts.
Step by step solution
01
Identify the coefficients
The given equation is in the form of a quadratic equation: \( y = ax^2 + bx + c \). Here, \( a = -1 \), \( b = 4 \), and \( c = -5 \). This is important for finding the vertex and axis of symmetry.
02
Determine the vertex
To find the vertex of the parabola, use the formula \( x = -\frac{b}{2a} \). Substituting \( b = 4 \) and \( a = -1 \), we get \( x = -\frac{4}{2(-1)} = 2 \). Plug \( x = 2 \) back into the equation to find \( y \): \( y = -(2)^2 + 4(2) - 5 = 4 - 5 = -1 \). Thus, the vertex is \( (2, -1) \).
03
Find the axis of symmetry
The axis of symmetry for a parabola in the form \( y = ax^2 + bx + c \) is the vertical line \( x = -\frac{b}{2a} \). From the calculation in Step 2, the axis of symmetry is \( x = 2 \).
04
Calculate the y-intercept
The y-intercept occurs where \( x = 0 \). Substituting \( x = 0 \) into the equation gives \( y = -(0)^2 + 4(0) - 5 = -5 \). So, the y-intercept is \( (0, -5) \).
05
Calculate the x-intercepts
To find the x-intercepts, set \( y = 0 \) and solve for \( x \). The equation becomes \( 0 = -x^2 + 4x - 5 \). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -1 \), \( b = 4 \), \( c = -5 \), we find \( x = \frac{-4 \pm \sqrt{16 - 20}}{-2} \). The discriminant \( 16 - 20 = -4 \) is negative, indicating there are no real x-intercepts.
06
Graph the parabola
Using a graphing tool or paper, plot the vertex \( (2, -1) \) and the y-intercept \( (0, -5) \). Draw the axis of symmetry as a vertical line through \( x = 2 \). Since there are no real x-intercepts, the parabola opens downward (because \( a = -1 \)) with the vertex as its highest point. Sketch the curvature from the vertex through the y-intercept.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
Quadratic equations are a vital aspect of algebra that deals with equations in the form of \( y = ax^2 + bx + c \). This type of equation is known as a polynomial equation of degree two, implying that the highest power of \( x \) is 2. In the context of graphing, this type of equation creates a curve known as a parabola. The way this parabola opens and its particular shape depends on the values of \( a \), \( b \), and \( c \).
- If \( a \) is positive, the parabola opens upwards, resembling a U-shape.
- If \( a \) is negative, it opens downwards, resembling an upside-down U-shape.
Vertex of a Parabola
The vertex of a parabola is a crucial point that represents the peak or the lowest point of the curve, depending on whether it opens upwards or downwards. To find the vertex, you can use the formula \[x = -\frac{b}{2a}\] which gives the x-coordinate of the vertex. In our specific equation \( y = -x^2 + 4x - 5 \), substituting \( a = -1 \) and \( b = 4 \) yields \[x = -\frac{4}{-2} = 2\].
Once you have the x-coordinate, plug it back into the original quadratic equation to find the y-coordinate. For this equation:
Once you have the x-coordinate, plug it back into the original quadratic equation to find the y-coordinate. For this equation:
- Substitute \( x = 2 \) back into the equation: \( y = -(2)^2 + 4(2) - 5 = -1 \)
- The vertex, therefore, is \( (2, -1) \).
Axis of Symmetry
The axis of symmetry is a line that divides a parabola into two mirrored halves. It is a critical property when graphing any quadratic equation. For a parabola given by the quadratic equation \( y = ax^2 + bx + c \), the axis of symmetry can be found using the same calculation as the x-coordinate of the vertex:\[x = -\frac{b}{2a}\]This vertical line runs through the vertex, ensuring that each point on one side of the parabola has a matching point directly opposite on the other side.
In the equation \( y = -x^2 + 4x - 5 \), using \( a = -1 \) and \( b = 4 \), the axis of symmetry is:
In the equation \( y = -x^2 + 4x - 5 \), using \( a = -1 \) and \( b = 4 \), the axis of symmetry is:
- \( x = 2 \)
Intercepts
Intercepts are points where the graph intersects the axes. These include the y-intercept and the potential x-intercepts.For the **y-intercept**:- It occurs where the graph crosses the y-axis (i.e., where \( x=0 \)).- Substitute \( x = 0 \) into the equation: \( y = -(0)^2 + 4(0) - 5 = -5 \).- So, the y-intercept for this equation is \( (0, -5) \).The **x-intercepts** are points where the graph crosses the x-axis (i.e., where \( y=0 \)). However, in this quadratic equation:\[ 0 = -x^2 + 4x - 5 \]Attempting to calculate the x-intercepts using the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]leads to a discriminant of \( 16 - 20 = -4 \), indicating that there are no real x-intercepts for this equation.Understanding intercepts aids in the graphing process and gives additional insights into the locations where the function interacts with the axes.