Problem 35
Question
Sketch the graph of the quadratic function. Identify the vertex and intercepts. $$f(x)=\frac{1}{4}\left(x^{2}-16 x+32\right)$$
Step-by-Step Solution
Verified Answer
The vertex of the function is at point (8,0). The function has x-intercepts at \(x = 8 \pm 4\sqrt{2} \approx 2.17\) and \(13.83\), and a y-intercept at \(y=8\). The graph is a parabola that opens upwards.
1Step 1: Identify the Vertex
The vertex of a quadratic function given in the form \(f(x) = a(x - h)^2 + k\) is at the point \((h, k)\). Rewriting the given function in this form will give the coordinates of the vertex. Doing so, we find that \(f(x) = \frac{1}{4}(x - 8)^2\), so the vertex is at \((8, 0)\).
2Step 2: Identify the Intercepts
The x-intercepts are the values of \(x\) for which \(f(x) = 0\). When we set \(f(x) = 0\), we find \(x = 8 \pm \sqrt{32}\), or \(x = 8 \pm 4\sqrt{2}\), approximately \(x= 2.17\) and \(x= 13.83\). For the y-intercept, replace \(x\) with 0 on the function, that gives us \(f(0) = 8\).
3Step 3: Sketch the Graph
Plot the vertex, intercepts and draw the curve of the graph. The vertex lies at (8,0) and the intercepts are at \(x=2.17\), \(x=13.83\), and \(y=8\). Since the value of \(a=\frac{1}{4}\) is positive, the graph opens upwards.
Key Concepts
Vertex of a Quadratic FunctionX-Intercepts and Y-InterceptsGraphing Quadratic Equations
Vertex of a Quadratic Function
Understanding the vertex of a quadratic function is crucial for sketching its graph. A quadratic function can be written in the form of \(f(x) = a(x - h)^2 + k\), where the point \(h, k\) is the vertex of the parabola. The vertex represents the highest or lowest point on the graph, depending on whether the parabola opens upwards or downwards.
Let's make this concept clearer with an example: suppose we have the function \(f(x)=\frac{1}{4}(x^2-16x+32)\). By completing the square, we rewrite it as \(f(x) = \frac{1}{4}(x - 8)^2\). In this form, it's easy to see that the vertex is \(8, 0\), which means it's the lowest point of the graph since \(\frac{1}{4}\) is positive and the parabola opens upwards. This information gives us a starting point for sketching the graph of the function and indicates where the curve will turn.
Let's make this concept clearer with an example: suppose we have the function \(f(x)=\frac{1}{4}(x^2-16x+32)\). By completing the square, we rewrite it as \(f(x) = \frac{1}{4}(x - 8)^2\). In this form, it's easy to see that the vertex is \(8, 0\), which means it's the lowest point of the graph since \(\frac{1}{4}\) is positive and the parabola opens upwards. This information gives us a starting point for sketching the graph of the function and indicates where the curve will turn.
X-Intercepts and Y-Intercepts
The points where a graph intersects the x-axis and y-axis are known as x-intercepts and y-intercepts, respectively. To find the x-intercepts of a quadratic function, we set the function equal to zero and solve for \(x\). In the case of our example function \(f(x)=\frac{1}{4}(x^2-16x+32)\), we get two solutions for the x-intercepts, which are approximately \(x=2.17\) and \(x=13.83\).
For the y-intercept, we simply plug in \(x=0\) into the function. This gives us the point where the parabola crosses the y-axis. In our example, we find \(f(0)=8\), which means the y-intercept is at the point \(0, 8\). These intercepts are essential for plotting the graph as they provide key points through which the parabola passes. By knowing where the graph crosses the axes, we can more accurately draw the curve of the quadratic function.
For the y-intercept, we simply plug in \(x=0\) into the function. This gives us the point where the parabola crosses the y-axis. In our example, we find \(f(0)=8\), which means the y-intercept is at the point \(0, 8\). These intercepts are essential for plotting the graph as they provide key points through which the parabola passes. By knowing where the graph crosses the axes, we can more accurately draw the curve of the quadratic function.
Graphing Quadratic Equations
Graphing a quadratic equation is a step-by-step process that begins by identifying key features like the vertex and intercepts. Once you have these, you can sketch the general shape of the parabola. A quadratic equation will result in a graph that's a parabola, a symmetric curve that opens either upwards or downwards depending on the sign of the coefficient \(a\).
In our exercise example, after determining the vertex and intercepts, we plot them on a coordinate plane. Starting at the vertex \(8, 0\), we know the graph opens upwards since \(a=\frac{1}{4}\) is positive. The graph will pass through the x-intercepts \(2.17, 0\) and \(13.83, 0\), and the y-intercept \(0, 8\). With these points, we can draw a smooth curve to represent the parabola, ensuring that it's symmetrical about the vertical line through the vertex, which in this case is the line \(x=8\). Graphing quadratic equations does not have to be complex; with practice and understanding of these key features, it becomes a manageable and systematic task.
In our exercise example, after determining the vertex and intercepts, we plot them on a coordinate plane. Starting at the vertex \(8, 0\), we know the graph opens upwards since \(a=\frac{1}{4}\) is positive. The graph will pass through the x-intercepts \(2.17, 0\) and \(13.83, 0\), and the y-intercept \(0, 8\). With these points, we can draw a smooth curve to represent the parabola, ensuring that it's symmetrical about the vertical line through the vertex, which in this case is the line \(x=8\). Graphing quadratic equations does not have to be complex; with practice and understanding of these key features, it becomes a manageable and systematic task.
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