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Magazine Chapter 3: Problem 15
Graph the given functions. $$y=6-2 x^{2}$$
Short Answer
Expert verified
The graph is a downward-opening parabola with vertex at (0, 6) and x-intercepts at (卤鈭�3, 0).
Step by step solution
01
Identify the Type of Function
The function given is a quadratic function of the form \( y = ax^2 + bx + c \). For the function \( y = 6 - 2x^2 \), we can see that \( a = -2 \), \( b = 0 \), and \( c = 6 \). Quadratic functions form a parabola when graphed.
02
Determine the Direction of the Parabola
The coefficient of \( x^2 \) is \( a = -2 \), which is negative. This means the parabola opens downwards.
03
Find the Vertex of the Parabola
For a quadratic function in the form \( y = ax^2 + bx + c \), the x-coordinate of the vertex is given by \( x = -\frac{b}{2a} \). Here, \( b = 0 \) and \( a = -2 \), so \( x = -\frac{0}{2(-2)} = 0 \). Substitute \( x = 0 \) into the function to find the y-coordinate: \( y = 6 - 2(0)^2 = 6 \). Therefore, the vertex is \((0, 6)\).
04
Find the Y-intercept
The y-intercept occurs where \( x = 0 \). We've already found this as the vertex point, which is \((0, 6)\).
05
Find the X-intercepts
To find the x-intercepts, solve \( 6 - 2x^2 = 0 \) for \( x \). Rearrange to \( 2x^2 = 6 \), then divide both sides by 2: \( x^2 = 3 \). Take the square root of both sides to get \( x = \pm \sqrt{3} \). Thus, the x-intercepts are \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\).
06
Sketch the Graph
Plot the vertex \((0, 6)\) and the intercepts \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\). Draw a symmetric parabolic curve opening downward, making sure it goes through these points. The vertex is the highest point on the graph because the parabola opens downward.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parabola
A quadratic function like the one provided, \(y=6-2x^2\), creates a curve known as a parabola when graphed. Parabolas can open up or down, or even sideways, but in this case, with the equation in the standard form \(y=ax^2+bx+c\), the parabola will open either upwards or downwards.
- If the coefficient \(a\) is positive, the parabola opens upwards, creating a U-shape.
- If \(a\) is negative, as with \(a=-2\) in our function, the parabola opens downwards, forming an upside-down U-shape.
Vertex
The vertex is a crucial element of a parabola, as it represents the maximum or minimum point. In this function \(y=6-2x^2\), the vertex is the peak because the parabola opens downward.
For quadratic functions of the form \(y=ax^2+bx+c\), you can find the x-coordinate of the vertex by using the formula \(x=-\frac{b}{2a}\). Given \(b=0\) and \(a=-2\), substituting these in gives us \(x=0\).
To find the y-coordinate, substitute \(x=0\) back into the original function, which results in \(y=6\).
Thus, the vertex is at the point \((0,6)\). This point is the highest on the graph and is symmetric about the y-axis.
For quadratic functions of the form \(y=ax^2+bx+c\), you can find the x-coordinate of the vertex by using the formula \(x=-\frac{b}{2a}\). Given \(b=0\) and \(a=-2\), substituting these in gives us \(x=0\).
To find the y-coordinate, substitute \(x=0\) back into the original function, which results in \(y=6\).
Thus, the vertex is at the point \((0,6)\). This point is the highest on the graph and is symmetric about the y-axis.
X-intercept
X-intercepts are points where the graph crosses the x-axis, meaning where \(y=0\). To find these for the function \(y=6-2x^2\), you need to solve for when the equation equals zero:
1. Set \(6 - 2x^2 = 0\).
2. Rearrange to get \(2x^2 = 6\).
3. Then \(x^2 = 3\) by dividing both sides by 2.
4. Solving \(x^2 = 3\) gives \(x=\pm\sqrt{3}\).
This means the x-intercepts are \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\). In other words, our parabola crosses the x-axis at these points.
1. Set \(6 - 2x^2 = 0\).
2. Rearrange to get \(2x^2 = 6\).
3. Then \(x^2 = 3\) by dividing both sides by 2.
4. Solving \(x^2 = 3\) gives \(x=\pm\sqrt{3}\).
This means the x-intercepts are \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\). In other words, our parabola crosses the x-axis at these points.
Y-intercept
The y-intercept is where the graph intersects the y-axis. At this point, \(x=0\).
To find the y-intercept of the function \(y=6-2x^2\), substitute \(x=0\) into the equation:
- Calculating gives \(y= 6 - 2(0)^2 = 6\).
Thus, the y-intercept is at the point \((0, 6)\). In this specific quadratic, the y-intercept is also the vertex, which is unique as the vertex is both the highest point and the point where the graph crosses the y-axis.
Understanding intercepts helps in visualizing the entire graph as these inform about where the parabola will touch and begin its curve from the axes.
To find the y-intercept of the function \(y=6-2x^2\), substitute \(x=0\) into the equation:
- Calculating gives \(y= 6 - 2(0)^2 = 6\).
Thus, the y-intercept is at the point \((0, 6)\). In this specific quadratic, the y-intercept is also the vertex, which is unique as the vertex is both the highest point and the point where the graph crosses the y-axis.
Understanding intercepts helps in visualizing the entire graph as these inform about where the parabola will touch and begin its curve from the axes.
