Chapter 3: Problem 41
Identify the vertex, axis of symmetry, y-intercept, x-intercepts, and opening of each parabola, then sketch the graph. $$y=x^{2}-3$$
Short Answer
Expert verified
Vertex: (0, -3), Axis of Symmetry: x = 0, y-intercept: (0, -3), x-intercepts: (√3, 0) and (-√3, 0), Opening: upward.
Step by step solution
01
- Identify the Vertex
For a quadratic equation in the form of \( y = ax^2 + bx + c \),the vertex \((h, k)\) can be found using the formula: \( h = -\frac{b}{2a} \). Here, \(a = 1\), \(b = 0\), and \(c = -3\). Substituting in,\( h = -\frac{0}{2*1} = 0 \). Substituting \(x = 0\) into the equation, \( y = 0^2 - 3 = -3 \). Therefore, the vertex is \( (0, -3) \).
02
- Find Axis of Symmetry
The axis of symmetry is the vertical line that passes through the vertex. It can be determined using \( x = h \). From the previous step, we found \( h = 0 \). Therefore, the axis of symmetry is \( x = 0 \).
03
- Determine the y-intercept
The y-intercept is the value of \( y \) when \( x = 0 \). Substituting \( x = 0 \) into the equation, \( y = 0^2 - 3 = -3 \). Therefore, the y-intercept is \( (0, -3) \).
04
- Find x-intercepts
The x-intercepts are the values of \( x \) for which \( y = 0 \). Setting \( y = 0 \) in the equation, \( 0 = x^2 - 3 \), solving for \( x \), we get \( x^2 = 3 \) or \( x = \pm\sqrt{3} \). Therefore, the x-intercepts are \( (\sqrt{3}, 0) \) and \((-\sqrt{3}, 0)\).
05
- Determine the Opening Direction
Since the coefficient of \( x^2 \) is positive \( (a = 1) \), the parabola opens upwards.
06
- Sketch the Graph
To sketch the graph, plot the vertex \( (0, -3) \), axis of symmetry \( x = 0 \), y-intercept \( (0, -3) \), and x-intercepts \((\sqrt{3}, 0) \) and \((-\sqrt{3}, 0) \). Since the parabola opens upwards, draw a smooth curve passing through these points.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
vertex
The vertex of a parabola is the highest or lowest point on its graph, depending on the direction the parabola opens. For a quadratic equation in the form of \( y = ax^2 + bx + c \), the vertex \((h, k)\) can be found using the formula: \( h = -\frac{b}{2a} \). Once you find \( h \), substitute it back into the equation to find \( k \).
In our example, \( y = x^2 - 3 \), we identified that \( a = 1 \), \( b = 0 \), and \( c = -3 \).
The calculations showed that \( h = 0 \), and substituting \( x = 0 \) back into the equation, we found \( y = -3 \).
Thus, the vertex is at \( (0, -3) \).
In our example, \( y = x^2 - 3 \), we identified that \( a = 1 \), \( b = 0 \), and \( c = -3 \).
The calculations showed that \( h = 0 \), and substituting \( x = 0 \) back into the equation, we found \( y = -3 \).
Thus, the vertex is at \( (0, -3) \).
axis of symmetry
The axis of symmetry is a vertical line that runs through the vertex of the parabola, dividing it into two symmetrical halves. This line can be determined using the equation \( x = h \).
For our quadratic equation \( y = x^2 - 3 \), we found that \( h = 0 \).
Therefore, the axis of symmetry is the line \( x = 0 \), meaning the parabola is symmetric about the y-axis.
For our quadratic equation \( y = x^2 - 3 \), we found that \( h = 0 \).
Therefore, the axis of symmetry is the line \( x = 0 \), meaning the parabola is symmetric about the y-axis.
y-intercept
The y-intercept is the point where the graph of the parabola crosses the y-axis. This happens when \( x = 0 \).
To find the y-intercept, substitute \( x = 0 \) into the quadratic equation. In our example, substituting \( x = 0 \) into \( y = x^2 - 3 \) gives \( y = -3 \).
Hence, the y-intercept is \( (0, -3) \). This is the same point as the vertex in this particular case.
To find the y-intercept, substitute \( x = 0 \) into the quadratic equation. In our example, substituting \( x = 0 \) into \( y = x^2 - 3 \) gives \( y = -3 \).
Hence, the y-intercept is \( (0, -3) \). This is the same point as the vertex in this particular case.
x-intercepts
The x-intercepts are the points where the graph of the parabola crosses the x-axis. These occur when \( y = 0 \).
To find the x-intercepts, set \( y = 0 \) in the quadratic equation and solve for \( x \). In our equation \( 0 = x^2 - 3 \), this simplifies to \( x^2 = 3 \), yielding \( x = \pm\sqrt{3} \).
Therefore, the x-intercepts are \( (\sqrt{3}, 0) \) and \( (-\sqrt{3}, 0) \). These points show where the parabola meets the x-axis.
To find the x-intercepts, set \( y = 0 \) in the quadratic equation and solve for \( x \). In our equation \( 0 = x^2 - 3 \), this simplifies to \( x^2 = 3 \), yielding \( x = \pm\sqrt{3} \).
Therefore, the x-intercepts are \( (\sqrt{3}, 0) \) and \( (-\sqrt{3}, 0) \). These points show where the parabola meets the x-axis.
parabola direction
The direction in which a parabola opens (upwards or downwards) is determined by the coefficient of \( x^2 \) (denoted as \( a \)) in the quadratic equation.
If \( a > 0 \), the parabola opens upwards.
If \( a < 0 \), the parabola opens downwards.
In our example, \( y = x^2 - 3 \), since \( a = 1 \) (a positive number), the parabola opens upwards. This tells us that the vertex is the lowest point on the graph, and the graph will extend upwards from this point.
If \( a > 0 \), the parabola opens upwards.
If \( a < 0 \), the parabola opens downwards.
In our example, \( y = x^2 - 3 \), since \( a = 1 \) (a positive number), the parabola opens upwards. This tells us that the vertex is the lowest point on the graph, and the graph will extend upwards from this point.