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Magazine Chapter 7: Problem 12
Sketch the graph of each parabola by using the vertex, the \(y\) -intercept, and the \(x\) -intercepts. Check the graph using a calculator. $$u=-3 v^{2}+12 v-9$$
Short Answer
Expert verified
The vertex is \((2, 3)\), the \(y\)-intercept is \((0, -9)\), and the \(x\)-intercepts are \((3, 0)\) and \((1, 0)\).
Step by step solution
01
Identify the Parabola Equation
The given equation is in the form \( u = -3v^2 + 12v - 9 \). This equation is a quadratic in terms of \( v \) and can be rewritten in the standard quadratic form \( av^2 + bv + c \), where \( a = -3 \), \( b = 12 \), and \( c = -9 \).
02
Find the Vertex
The vertex \( (h, k) \) of a parabola given by \( av^2 + bv + c \) can be found using the formulas \( h = -\frac{b}{2a} \) and calculating \( k \) by substituting \( h \) for \( v \) in the equation \( u = -3v^2 + 12v - 9 \).Calculate \( h = -\frac{12}{2(-3)} = 2 \).Substitute \( v = 2 \) into the equation:\( k = -3(2)^2 + 12(2) - 9 = -3(4) + 24 - 9 = -12 + 24 - 9 = 3 \\).So, the vertex is \( (2, 3) \).
03
Find the y-intercept
The \( y \)-intercept occurs when \( v = 0 \). Substitute \( v = 0 \) into the equation to find \( u \):\( u = -3(0)^2 + 12(0) - 9 = -9 \).So, the \( y \)-intercept is \( (0, -9) \).
04
Find the x-intercepts
The \( x \)-intercepts occur where \( u = 0 \). Set the equation equal to zero and solve:\( 0 = -3v^2 + 12v - 9 \).Divide every term by \(-3\):\( 0 = v^2 - 4v + 3 \).Factor the quadratic:\( (v - 3)(v - 1) = 0 \).So, \( v = 3 \) and \( v = 1 \). The \( x \)-intercepts are \( (3, 0) \) and \( (1, 0) \).
05
Sketch the Graph
Plot the vertex \( (2, 3) \), the \( y \)-intercept \( (0, -9) \), and the \( x \)-intercepts \( (3, 0) \) and \( (1, 0) \) on a graph. Connect these points to form a parabola. The parabola opens downwards since the coefficient of \( v^2 \) is negative.
06
Verify with a Calculator
Use a graphing calculator to enter the equation \( u = -3v^2 + 12v - 9 \). Confirm that the calculator graph matches your sketch, including the vertex, \( y \)-intercept, and \( x \)-intercepts.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertex Calculation
The vertex is a critical point of a parabola where it changes direction. It provides valuable information about the parabola's maximum or minimum point. In our equation, \( u = -3v^2 + 12v - 9 \), we find the vertex using the formula for the vertex \( (h, k) \). To calculate \( h \), use the formula \( h = -\frac{b}{2a} \). In this case, \( a = -3 \) and \( b = 12 \), so \( h = -\frac{12}{2 \times -3} = 2 \).
Once you have \( h \), substitute it back into the equation to find \( k \): \( k = -3(2)^2 + 12 \cdot 2 - 9 = 3 \).
Therefore, the vertex is \( (2, 3) \). This vertex is essential for sketching the graph and understanding the parabola's properties.
Once you have \( h \), substitute it back into the equation to find \( k \): \( k = -3(2)^2 + 12 \cdot 2 - 9 = 3 \).
Therefore, the vertex is \( (2, 3) \). This vertex is essential for sketching the graph and understanding the parabola's properties.
Y-Intercept
The \( y \)-intercept is where the parabola crosses the \( y \)-axis. This occurs when \( v = 0 \). To find it, simply substitute \( v = 0 \) into the equation \( u = -3v^2 + 12v - 9 \).
Calculate the value of \( u \) when \( v = 0 \):
The \( y \)-intercept shows where the graph starts from on the \( y \)-axis and is part of the initial sketch of the graph.
Calculate the value of \( u \) when \( v = 0 \):
- \( u = -3 \times 0^2 + 12 \times 0 - 9 = -9 \)
The \( y \)-intercept shows where the graph starts from on the \( y \)-axis and is part of the initial sketch of the graph.
X-Intercepts
Finding the \( x \)-intercepts involves determining where the parabola crosses the \( x \)-axis, which means setting \( u = 0 \). For our equation, this means solving: \(-3v^2 + 12v - 9 = 0 \).
To simplify:
These intercepts show where the parabola meets the \( x \)-axis.
To simplify:
- Divide all terms by \(-3\) to make it more manageable: \( v^2 - 4v + 3 = 0 \)
- Factor the quadratic to find: \((v - 3)(v - 1) = 0 \)
These intercepts show where the parabola meets the \( x \)-axis.
Quadratic Equation
A quadratic equation is of the form \( av^2 + bv + c = 0 \). It's essential in graphing parabolas because it defines the curve. In our problem, the quadratic equation is \(-3v^2 + 12v - 9 \).
Here:
Understanding these elements helps in vertex calculation, symmetry determination, and sketching the parabola accurately.
Here:
- \( a = -3 \) is the coefficient of \( v^2 \)
- \( b = 12 \) is the coefficient of \( v \)
- \( c = -9 \) is the constant term
Understanding these elements helps in vertex calculation, symmetry determination, and sketching the parabola accurately.
Graphing Calculator
A graphing calculator is a powerful tool to verify your parabola sketches. After computing the vertex and intercepts, input the equation \( u = -3v^2 + 12v - 9 \) into the calculator.
This will produce a graph that should display:
It complements your mathematical calculations by showing the actual shape and position of the parabola on a graph.
This will produce a graph that should display:
- The vertex at \( (2, 3) \)
- The \( y \)-intercept at \( (0, -9) \)
- The \( x \)-intercepts at \( (3, 0) \) and \( (1, 0) \)
It complements your mathematical calculations by showing the actual shape and position of the parabola on a graph.
