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Chapter 2: Problem 9

Find the coordinates of the vertex for the parabola defined by the given quadratic function. $$f(x)=2(x-3)^{2}+1$$

Short Answer

Expert verified
The vertex of the parabola defined by the function \(f(x)=2(x-3)^{2}+1\) is at (3,1).

Step by step solution

01

Identify the Vertex Form

The given equation is in the form \(f(x) = a(x-h)^2 + k\). In this form, the coordinates of the vertex of the parabola are given by the values of \(h\) and \(k\). Here, a=2, h=3, and k=1.
02

Determine the Vertex

From the equation \(f(x)=2(x-3)^{2}+1\), we can see that h=3, k=1. Hence, the vertex of the given parabola is at the coordinates \((h,k)\), which are \((3,1)\).

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Most popular questions from this chapter

Use a graphing utility to graph $$f(x)=\frac{x^{2}-4 x+3}{x-2} \text { and } g(x)=\frac{x^{2}-5 x+6}{x-2}$$ What differences do you observe between the graph of \(f\) and the graph of \(g\) ? How do you account for these differences?

Use inspection to describe each inequality's solution set. Do not solve any of the inequalities. $$(x-2)^{2} \leq 0$$

Use a graphing utility to graph \(y=\frac{1}{x}, y=\frac{1}{x^{3}},\) and \(\frac{1}{x^{5}}\) in the same viewing rectangle. For odd values of \(n,\) how does changing \(n\) affect the graph of \(y=\frac{1}{x^{n}} ?\)

Use transformations of \(f(x)=\frac{1}{x}\) or \(f(x)=\frac{1}{x^{2}}\) to graph each rational function. $$g(x)=\frac{1}{x-1}$$

a. Find the slant asymptote of the graph of each rational function and \(\mathbf{b} .\) Follow the seven-step strategy and use the slant asymptote to graph each rational function. $$f(x)=\frac{x^{2}-4}{x}$$
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