Chapter 5: Problem 3
Graph each function. $$ y=-2 x^{2}+16 x-31 $$
Short Answer
Expert verified
Graph the parabola with vertex at (4, 1), opening downward, and passing through points (0, -31), (3, -1), and (5, -1).
Step by step solution
01
Identify the type of function
The given function is in the form of \(y = ax^2 + bx + c\), which is a quadratic function. Quadratic functions form a parabola when graphed.
02
Find the vertex
For the quadratic equation \(y = ax^2 + bx + c\), the vertex form is found using \(x = -\frac{b}{2a}\). Here, \(a = -2\), \(b = 16\). Applying the formula: \(x = -\frac{16}{2(-2)} = 4\). Substitute \(x = 4\) back into the function to find \(y\): \(y = -2(4)^2 + 16(4) - 31 = 1\). So, the vertex is \((4, 1)\).
03
Identify the axis of symmetry
The axis of symmetry for a parabola described by \(y = ax^2 + bx + c\) is the vertical line that passes through the vertex, so the axis of symmetry is \(x = 4\).
04
Determine the direction of the parabola
Since \(a = -2\) and it's negative, the parabola opens downward.
05
Find the y-intercept
The y-intercept occurs where \(x = 0\). Substitute \(x = 0\) into the function: \(y = -2(0)^2 + 16(0) - 31 = -31\). So the y-intercept is \((0, -31)\).
06
Find additional points
Choose another value for \(x\) to find more points on the parabola. Let \(x = 5\): \(y = -2(5)^2 + 16(5) - 31 = -1\), so another point is \((5, -1)\). Let \(x = 3\): \(y = -2(3)^2 + 16(3) - 31 = -1\), so another point is \((3, -1)\).
07
Plot the points and sketch the graph
Plot the vertex \((4, 1)\), the axis of symmetry \(x = 4\), the y-intercept \((0, -31)\), and the additional points \((3, -1)\) and \((5, -1)\). Sketch the parabola passing through these points, opening downward.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Vertex
The vertex of a quadratic function gives you a central point on its graph, usually forming the highest or lowest point on the curve. For a parabola, this is where the curve turns. Knowing the vertex is key because it helps to easily sketch the graph of the function.
To find the vertex of a quadratic function in the form of \(y = ax^2 + bx + c\), use the formula \(x = -\frac{b}{2a}\).
Once you have \(x\), substitute it back into the equation to find the \(y\)-value. For example, for the quadratic equation \(y = -2x^2 + 16x - 31\), we calculated the vertex at \((4, 1)\).
This reveals that the curve reaches its peak at this point, as the coefficient \(a\) is negative, indicating a parabola that opens downward.
To find the vertex of a quadratic function in the form of \(y = ax^2 + bx + c\), use the formula \(x = -\frac{b}{2a}\).
Once you have \(x\), substitute it back into the equation to find the \(y\)-value. For example, for the quadratic equation \(y = -2x^2 + 16x - 31\), we calculated the vertex at \((4, 1)\).
This reveals that the curve reaches its peak at this point, as the coefficient \(a\) is negative, indicating a parabola that opens downward.
The Axis of Symmetry
The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. It always passes through the vertex of the parabola.
For the quadratic function \(y = ax^2 + bx + c\), the axis of symmetry is given by the formula \(x = -\frac{b}{2a}\), which means it shares the \(x\)-coordinate with the vertex.
For the quadratic function \(y = ax^2 + bx + c\), the axis of symmetry is given by the formula \(x = -\frac{b}{2a}\), which means it shares the \(x\)-coordinate with the vertex.
- It's an important feature because it helps to predict the behavior of the graph.
- For example, any two points on opposite sides of the axis will be equidistant from it.
Locating the Y-Intercept
The \(y\)-intercept of a quadratic function is the point where the parabola crosses the \(y\)-axis. This happens when \(x = 0\) in the equation. To find this intercept, simply set \(x = 0\) and solve for \(y\).
For the quadratic function \(y = -2x^2 + 16x - 31\), substituting \(x = 0\) yields \(y = -31\), meaning the \(y\)-intercept is \((0, -31)\).
For the quadratic function \(y = -2x^2 + 16x - 31\), substituting \(x = 0\) yields \(y = -31\), meaning the \(y\)-intercept is \((0, -31)\).
- Knowing the \(y\)-intercept gives you a starting point for graphing the curve.
- It's essential for understanding where the parabola sits relative to the \(y\)-axis.
Properties of a Parabola
A parabola is the symmetric curve formed by a quadratic function. It can open upwards or downwards depending on the sign of \(a\) in the equation \(y = ax^2 + bx + c\).
In our example, since \(a = -2\), the parabola opens downward.
In our example, since \(a = -2\), the parabola opens downward.
- If \(a\) is positive, the parabola opens upward, resembling a cup.
- A negative \(a\) makes it open downward like an umbrella.
Steps for Graphing Quadratic Equations
Graphing quadratic equations involves a few straightforward steps. Start by determining the vertex using the formula \(x = -\frac{b}{2a}\) and calculating the corresponding \(y\)-value.
- Find the axis of symmetry, which passes through the vertex vertically.
- Calculate the \(y\)-intercept by setting \(x = 0\).
- Recognize whether the parabola opens upwards or downwards based on the sign of \(a\).