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Q. 25

Expert-verified
Found in: Page 156

Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

Graph the function $f$ by starting with the graph of $y={x}^{2}$ and using transformations (shifting, compressing, stretching, and/or reflection). Verify your results using a graphing utility.Hint: If necessary, write $f$ in the form $f\left(x\right)=a\left(x-h{\right)}^{2}+k$.$f\left(x\right)=2{x}^{2}-4x+1$

The required graph is shown below:

See the step by step solution

Step 1. Write the given function in vertex form.

The given function is:

$f\left(x\right)=2{x}^{2}-4x+1\phantom{\rule{0ex}{0ex}}f\left(x\right)=2\left({x}^{2}-2x\right)+1$

Add and subtract the square of half of the coefficient of x inside the parenthesis.

$f\left(x\right)=2\left({x}^{2}-2x+1-1\right)+1\phantom{\rule{0ex}{0ex}}f\left(x\right)=2\left({x}^{2}-2x+1\right)-2+1\phantom{\rule{0ex}{0ex}}f\left(x\right)=2\left(x-1{\right)}^{2}-1$

Step 2. Determine the transformations used.

In the function $f\left(x\right)=a\left(x-h{\right)}^{2}+k$, a is a constant and $\left(h,k\right)$ is the vertex.

In the given function $a=2,h=1,k=-1$. It means the graph of the given function is a parabola that opens up and has its vertex at $\left(1,-1\right)$ and its axis of symmetry is the line $x=1$.

The graph of $y={x}^{2}$ vertically stretched by factor 2, shifts 1 units right and 1 units down.

First, plot the graph of $y={x}^{2}$ then multiply the values by 2 to get the graph of $y=2{x}^{2}$, then shift it 1 unit right to get the graph of the function localid="1646042810491" $y=2\left(x-1{\right)}^{2}$. After that shift the resulting graph 1 unit down to get the graph of the function $f\left(x\right)=2\left(x-1{\right)}^{2}-1$.

Step 3. Draw the graph.

The required graph is shown below:

Using a graphing utility, we get a graph that is the same as the above graph.