{"id":2556,"date":"2025-08-13T18:03:57","date_gmt":"2025-08-13T18:03:57","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=2556"},"modified":"2026-01-12T19:13:21","modified_gmt":"2026-01-12T19:13:21","slug":"polynomial-functions-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/polynomial-functions-background-youll-need-1\/","title":{"raw":"Polynomial Functions: Background You'll Need 1","rendered":"Polynomial Functions: Background You&#8217;ll Need 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Simplify radical expressions<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Simplifying Square Roots and Expressing Them in Lowest Terms<\/h2>\r\nTo <strong>simplify a square root<\/strong> means that we rewrite the square root as a rational number times the square root of a number that has no perfect square factors. The act of changing a square root into such a form is simplifying the square root. Before discussing how to simplify a square root, we need to introduce a rule about square roots.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>the product rule for square roots<\/h3>\r\nThe square root of a product of numbers equals the product of the square roots of those number.\r\n\r\nGiven that [latex]a[\/latex] and [latex]b[\/latex] are nonnegative real numbers,\r\n\r\n<center>[latex]\\sqrt{a \\times {b}}=\\sqrt{a} \\times \\sqrt{b}[\/latex]<\/center><\/div>\r\n<\/section>Using this formula, we can factor an integer inside a square root into a perfect square times another integer. Then the square root can be applied to the perfect square, leaving an integer times the square root of another integer. If the number remaining under the square root has no perfect square factors, then we\u2019ve simplified the irrational number into lowest terms.\r\n\r\n<section class=\"textbox proTip\">A perfect square is an integer that can be expressed as the square of another integer. For example, [latex]16[\/latex], [latex]25[\/latex], and [latex]36[\/latex] are perfect squares because they are [latex]4^2[\/latex], [latex]5^2[\/latex], and [latex]6^2[\/latex], respectively.<\/section><section class=\"textbox questionHelp\"><strong>How to: Simplify square roots into lowest terms when [latex]n[\/latex] is an integer<\/strong>\r\n<ul>\r\n \t<li><strong>Step 1:<\/strong> Determine the largest perfect square factor of [latex]n[\/latex], which we denote [latex]a^2[\/latex].<\/li>\r\n \t<li><strong>Step 2:<\/strong> Factor [latex]n[\/latex] into [latex]a^2\u00d7b[\/latex].<\/li>\r\n \t<li><strong>Step 3:<\/strong> Apply [latex]\\sqrt{a^2 \\times b} =\\sqrt{a^2} \\times \\sqrt{b}[\/latex].<\/li>\r\n \t<li><strong>Step 4:<\/strong> Write [latex]\\sqrt{n}[\/latex] in its simplified form, [latex]a\\sqrt{b}[\/latex].<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\">To simplify the given radical expressions, we'll break down the numbers into their [pb_glossary id=\"662\"]prime factors[\/pb_glossary] and simplify the radicals accordingly, while also considering the powers of the variables. Here are the steps:First, factor [latex]300[\/latex] into its prime factors:\r\n<p style=\"text-align: center;\">[latex]300 = 2^2 \\cdot 3 \\cdot 5^2[\/latex]<\/p>\r\nNow, extract the square roots of the perfect squares:\r\n\r\n[latex]\\begin{align} \\sqrt{300} &amp;= \\sqrt{2^2 \\cdot 3 \\cdot 5^2} &amp;&amp; \\text{Factor the number into prime factors.} \\\\ &amp;= \\sqrt{2^2} \\cdot \\sqrt{3} \\cdot \\sqrt{5^2} &amp;&amp; \\text{Separate each factor under its own square root.} \\\\ &amp;= 2 \\cdot \\sqrt{3} \\cdot 5 &amp;&amp; \\text{Simplify the square roots of perfect squares.} \\\\ &amp;= 10\\sqrt{3} &amp;&amp; \\text{Multiply the results to get the simplified form.} \\end{align}[\/latex]\r\n\r\n<\/section><section class=\"textbox example\">Simplify [latex]\\sqrt{162{a}^{5}{b}^{4}}[\/latex].[reveal-answer q=\"30444\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"30444\"]<span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">\\begin{align} \\sqrt{162a^5b^4} &amp;= \\sqrt{2 \\cdot 3^4 \\cdot a^5 \\cdot b^4} &amp;&amp; \\text{Factor the number into prime factors and express variables.} \\\\ &amp;= \\sqrt{2 \\cdot (3^2)^2 \\cdot a^4 \\cdot a \\cdot (b^2)^2} &amp;&amp; \\text{Break down the expression to show squares for clarity.} \\\\ &amp;= \\sqrt{2} \\cdot \\sqrt{(3^2)^2} \\cdot \\sqrt{a^4} \\cdot \\sqrt{a} \\cdot \\sqrt{(b^2)^2} &amp;&amp; \\text{Separate each factor under its own square root.} \\\\ &amp;= \\sqrt{2} \\cdot 3^2 \\cdot a^2 \\cdot \\sqrt{a} \\cdot b^2 &amp;&amp; \\text{Simplify the square roots of perfect squares.} \\\\ &amp;= 9a^2b^2 \\cdot \\sqrt{2a} &amp;&amp; \\text{Combine the constants and simplify further to finalize.} \\end{align}[\/hidden-answer]<\/span><\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]318776[\/ohm_question]<\/section><section class=\"textbox proTip\">For the variable [latex]x[\/latex], [latex]\\sqrt{x^2} = |x|[\/latex] , but why is that?When you square any values, the result is always non-negative, meaning it's either positive or zero. Then, when you take the square root of this non-negative squared value, you get back the original number without its sign\u2014just its size or magnitude. Thus, taking the square root of [latex]x^2[\/latex] always yields the absolute value of [latex]x[\/latex] ensuring that we consider [latex]x[\/latex] in its non-negative form.<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]318777[\/ohm_question]<\/section>Given the product of multiple radical expressions, we can use the product rule to combine them into one radical expression and then simplify as we did above.\r\n\r\n<section class=\"textbox example\">Simplify the following radical expression.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"text-align: left;\">[latex]\\sqrt{12}\\cdot \\sqrt{3}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{50x}\\cdot \\sqrt{2x}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"13428714\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"13428714\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\begin{align}\\sqrt{12}\\cdot \\sqrt{3} &amp; = \\sqrt{12\\cdot 3} &amp;&amp; \\text{Express the product as a single radical expression}. \\\\ &amp;= \\sqrt{36} &amp;&amp; \\text{Simplify}. \\\\ &amp;= 6 \\end{align}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{align*} \\sqrt{50x} \\cdot \\sqrt{2x} &amp;= \\sqrt{50x \\cdot 2x} &amp; \\text{Multiply under the radicals} \\\\ &amp;= \\sqrt{100x^2} &amp; \\text{Simplify the product inside the radical} \\\\ &amp;= 10|x| &amp; \\text{Take the square root of \\(100\\) and \\(x^2\\)} \\end{align*}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]318778[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Simplify radical expressions<\/li>\n<\/ul>\n<\/section>\n<h2>Simplifying Square Roots and Expressing Them in Lowest Terms<\/h2>\n<p>To <strong>simplify a square root<\/strong> means that we rewrite the square root as a rational number times the square root of a number that has no perfect square factors. The act of changing a square root into such a form is simplifying the square root. Before discussing how to simplify a square root, we need to introduce a rule about square roots.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>the product rule for square roots<\/h3>\n<p>The square root of a product of numbers equals the product of the square roots of those number.<\/p>\n<p>Given that [latex]a[\/latex] and [latex]b[\/latex] are nonnegative real numbers,<\/p>\n<div style=\"text-align: center;\">[latex]\\sqrt{a \\times {b}}=\\sqrt{a} \\times \\sqrt{b}[\/latex]<\/div>\n<\/div>\n<\/section>\n<p>Using this formula, we can factor an integer inside a square root into a perfect square times another integer. Then the square root can be applied to the perfect square, leaving an integer times the square root of another integer. If the number remaining under the square root has no perfect square factors, then we\u2019ve simplified the irrational number into lowest terms.<\/p>\n<section class=\"textbox proTip\">A perfect square is an integer that can be expressed as the square of another integer. For example, [latex]16[\/latex], [latex]25[\/latex], and [latex]36[\/latex] are perfect squares because they are [latex]4^2[\/latex], [latex]5^2[\/latex], and [latex]6^2[\/latex], respectively.<\/section>\n<section class=\"textbox questionHelp\"><strong>How to: Simplify square roots into lowest terms when [latex]n[\/latex] is an integer<\/strong><\/p>\n<ul>\n<li><strong>Step 1:<\/strong> Determine the largest perfect square factor of [latex]n[\/latex], which we denote [latex]a^2[\/latex].<\/li>\n<li><strong>Step 2:<\/strong> Factor [latex]n[\/latex] into [latex]a^2\u00d7b[\/latex].<\/li>\n<li><strong>Step 3:<\/strong> Apply [latex]\\sqrt{a^2 \\times b} =\\sqrt{a^2} \\times \\sqrt{b}[\/latex].<\/li>\n<li><strong>Step 4:<\/strong> Write [latex]\\sqrt{n}[\/latex] in its simplified form, [latex]a\\sqrt{b}[\/latex].<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\">To simplify the given radical expressions, we&#8217;ll break down the numbers into their <a class=\"glossary-term\" aria-haspopup=\"dialog\" aria-describedby=\"definition\" href=\"#term_2556_662\">prime factors<\/a> and simplify the radicals accordingly, while also considering the powers of the variables. Here are the steps:First, factor [latex]300[\/latex] into its prime factors:<\/p>\n<p style=\"text-align: center;\">[latex]300 = 2^2 \\cdot 3 \\cdot 5^2[\/latex]<\/p>\n<p>Now, extract the square roots of the perfect squares:<\/p>\n<p>[latex]\\begin{align} \\sqrt{300} &= \\sqrt{2^2 \\cdot 3 \\cdot 5^2} && \\text{Factor the number into prime factors.} \\\\ &= \\sqrt{2^2} \\cdot \\sqrt{3} \\cdot \\sqrt{5^2} && \\text{Separate each factor under its own square root.} \\\\ &= 2 \\cdot \\sqrt{3} \\cdot 5 && \\text{Simplify the square roots of perfect squares.} \\\\ &= 10\\sqrt{3} && \\text{Multiply the results to get the simplified form.} \\end{align}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">Simplify [latex]\\sqrt{162{a}^{5}{b}^{4}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q30444\">Show Answer<\/button><\/p>\n<div id=\"q30444\" class=\"hidden-answer\" style=\"display: none\"><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">\\begin{align} \\sqrt{162a^5b^4} &amp;= \\sqrt{2 \\cdot 3^4 \\cdot a^5 \\cdot b^4} &amp;&amp; \\text{Factor the number into prime factors and express variables.} \\\\ &amp;= \\sqrt{2 \\cdot (3^2)^2 \\cdot a^4 \\cdot a \\cdot (b^2)^2} &amp;&amp; \\text{Break down the expression to show squares for clarity.} \\\\ &amp;= \\sqrt{2} \\cdot \\sqrt{(3^2)^2} \\cdot \\sqrt{a^4} \\cdot \\sqrt{a} \\cdot \\sqrt{(b^2)^2} &amp;&amp; \\text{Separate each factor under its own square root.} \\\\ &amp;= \\sqrt{2} \\cdot 3^2 \\cdot a^2 \\cdot \\sqrt{a} \\cdot b^2 &amp;&amp; \\text{Simplify the square roots of perfect squares.} \\\\ &amp;= 9a^2b^2 \\cdot \\sqrt{2a} &amp;&amp; \\text{Combine the constants and simplify further to finalize.} \\end{align}<\/div>\n<\/div>\n<p><\/span><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm318776\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=318776&theme=lumen&iframe_resize_id=ohm318776&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox proTip\">For the variable [latex]x[\/latex], [latex]\\sqrt{x^2} = |x|[\/latex] , but why is that?When you square any values, the result is always non-negative, meaning it&#8217;s either positive or zero. Then, when you take the square root of this non-negative squared value, you get back the original number without its sign\u2014just its size or magnitude. Thus, taking the square root of [latex]x^2[\/latex] always yields the absolute value of [latex]x[\/latex] ensuring that we consider [latex]x[\/latex] in its non-negative form.<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm318777\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=318777&theme=lumen&iframe_resize_id=ohm318777&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<p>Given the product of multiple radical expressions, we can use the product rule to combine them into one radical expression and then simplify as we did above.<\/p>\n<section class=\"textbox example\">Simplify the following radical expression.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"text-align: left;\">[latex]\\sqrt{12}\\cdot \\sqrt{3}[\/latex]<\/li>\n<li>[latex]\\sqrt{50x}\\cdot \\sqrt{2x}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q13428714\">Show Solution<\/button><\/p>\n<div id=\"q13428714\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\begin{align}\\sqrt{12}\\cdot \\sqrt{3} & = \\sqrt{12\\cdot 3} && \\text{Express the product as a single radical expression}. \\\\ &= \\sqrt{36} && \\text{Simplify}. \\\\ &= 6 \\end{align}[\/latex]<\/li>\n<li>[latex]\\begin{align*} \\sqrt{50x} \\cdot \\sqrt{2x} &= \\sqrt{50x \\cdot 2x} & \\text{Multiply under the radicals} \\\\ &= \\sqrt{100x^2} & \\text{Simplify the product inside the radical} \\\\ &= 10|x| & \\text{Take the square root of \\(100\\) and \\(x^2\\)} \\end{align*}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm318778\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=318778&theme=lumen&iframe_resize_id=ohm318778&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<div class=\"glossary\"><span class=\"screen-reader-text\" id=\"definition\">definition<\/span><template id=\"term_2556_662\"><div class=\"glossary__definition\" role=\"dialog\" data-id=\"term_2556_662\"><div tabindex=\"-1\"><section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>transformed functions<\/h3>\n<p>The formula for a transformed function is [latex]g(x) = \\pm a \\cdot f\\big(\\pm b(x - h)\\big) + k[\/latex] where:<\/p>\n<ul>\n<li>[latex]\\pm a[\/latex] describes the vertical reflection and stretch\/compression<\/li>\n<li>[latex]\\pm b[\/latex] describes the horizontal reflection and stretch\/compression<\/li>\n<li>[latex]h[\/latex] describes the horizontal shift, and<\/li>\n<li>[latex]k[\/latex] describes the vertical shift<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<ol>\n<li>Find the parent function. If it's not given to you, check the toolkit functions.<\/li>\n<li>Identify any shifts.<\/li>\n<li>Identify any reflections, stretches or compresses.<\/li>\n<li>Write the function using [latex]g(x) = a \\cdot f\\big(b(x - h)\\big) + k[\/latex]<\/li>\n<\/ol>\n<\/section>\n<h3>Writing Functions Given the Transformed Graph<\/h3>\n<section class=\"textbox example\" aria-label=\"Example\"><img class=\"wp-image-1606 alignright\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/30012528\/Screenshot-2024-05-29-at-6.25.19%E2%80%AFPM.png\" alt=\"\" width=\"309\" height=\"301\" \/>The graph shows two function: The toolkit function [latex]f(x) = x^3[\/latex] (green) and [latex]g(x)[\/latex] (red). Relate this new function [latex]g\\left(x\\right)[\/latex] to [latex]f\\left(x\\right)[\/latex], and then find a formula for [latex]g\\left(x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q343479\">Show Answer<\/button><\/p>\n<div id=\"q343479\" class=\"hidden-answer\" style=\"display: none\">The red curve [latex]g(x)[\/latex] appears to be less steep compared to the green curve [latex]f(x)[\/latex]. This suggests a vertical compression.If [latex]g(x)[\/latex] is a vertical compression of [latex]f(x)[\/latex], we have: [latex]g(x) = a \\cdot f(x)[\/latex], where [latex]0 < a < 1[\/latex].To determine [latex]a[\/latex], it is helpful to look for a point on the graph that is relatively clear.\n\n\n<ul>\n<li>In this graph, it appears that [latex]g\\left(2\\right)=2[\/latex].<\/li>\n<li>With the basic cubic function at the same input, [latex]f\\left(2\\right)={2}^{3}=8[\/latex].<\/li>\n<li>Based on that, it appears that the outputs of [latex]g[\/latex] are [latex]\\frac{1}{4}[\/latex] the outputs of the function [latex]f[\/latex] because [latex]2=\\frac{1}{4} \\cdot 8[\/latex].<\/li>\n<\/ul>\n<p>Thus, [latex]g(x) = \\frac{1}{4} f(x) = \\frac{1}{4} x^3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\"><img class=\"alignright\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18203628\/CNX_Precalc_Figure_01_05_032.jpg\" alt=\"Graph of f(x) being vertically compressed to g(x).\" width=\"487\" height=\"291\" \/>Relate the function [latex]g\\left(x\\right)[\/latex] to [latex]f\\left(x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q623190\">Show Answer<\/button><\/p>\n<div id=\"q623190\" class=\"hidden-answer\" style=\"display: none\">The orange graph\u00a0 [latex]g(x)[\/latex] appears to be a horizontally compressed version of the blue graph of [latex]f(x)[\/latex].<br \/>\n[latex]\\\\[\/latex]<br \/>\nIf [latex]g(x)[\/latex] is a horizontal compression of [latex]f(x)[\/latex], we have: [latex]g(x) = f(b \\cdot x)[\/latex], where [latex]b > 1[\/latex]. The graph is compressed by [latex]\\dfrac{1}{b}[\/latex].To determine [latex]b[\/latex], it is helpful to look for a point on the graph that is relatively clear.<\/p>\n<ul>\n<li>In the compressed graph [latex]g(x)[\/latex], the end point is [latex](2, 4)[\/latex].<\/li>\n<li>The end point of [latex]f(x)[\/latex] is [latex](6,4)[\/latex].<\/li>\n<li>We can see that the [latex]x[\/latex]-values have been compressed by [latex]\\frac{1}{3}[\/latex], because [latex]2=\\frac{1}{3} \\cdot 6[\/latex].<\/li>\n<li>This means that [latex]\\dfrac{1}{b} = \\dfrac{1}{3}[\/latex], which means [latex]b = 3[\/latex].<\/li>\n<\/ul>\n<p>Thus, [latex]g(x)=f(3x)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">The graph below represents a transformation of the toolkit function [latex]f\\left(x\\right)={x}^{2}[\/latex]. Relate this new function [latex]g\\left(x\\right)[\/latex] to [latex]f\\left(x\\right)[\/latex], and then find a formula for [latex]g\\left(x\\right)[\/latex].<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18203554\/CNX_Precalc_Figure_01_05_0072.jpg\" alt=\"Graph of a parabola.\" width=\"487\" height=\"328\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q937293\">Show Solution<\/button><\/p>\n<div id=\"q937293\" class=\"hidden-answer\" style=\"display: none\">Notice that the graph is identical in shape to the [latex]f\\left(x\\right)={x}^{2}[\/latex] function, but the [latex]x[\/latex]<em>-<\/em>values are shifted to the right 2 units. The vertex used to be at (0,0), but now the vertex is at (2,0). The graph is the basic quadratic function shifted 2 units to the right, so<\/p>\n<p style=\"text-align: center;\">[latex]g\\left(x\\right)=f\\left(x - 2\\right)[\/latex]<\/p>\n<p>Notice how we must input the value [latex]x=2[\/latex] to get the output value [latex]y=0[\/latex]; the [latex]x[\/latex]-values must be 2 units larger because of the shift to the right by 2 units. We can then use the definition of the [latex]f\\left(x\\right)[\/latex] function to write a formula for [latex]g\\left(x\\right)[\/latex] by evaluating [latex]f\\left(x - 2\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}f\\left(x\\right)={x}^{2}\\hfill \\\\ g\\left(x\\right)=f\\left(x - 2\\right)\\hfill \\\\ g\\left(x\\right)=f\\left(x - 2\\right)={\\left(x - 2\\right)}^{2}\\hfill \\end{cases}[\/latex]<\/p>\n<p><strong>Analysis of the Solution<\/strong><\/p>\n<p>To determine whether the shift is [latex]+2[\/latex] or [latex]-2[\/latex] , consider a single reference point on the graph. For a quadratic, looking at the vertex point is convenient. In the original function, [latex]f\\left(0\\right)=0[\/latex]. In our shifted function, [latex]g\\left(2\\right)=0[\/latex]. To obtain the output value of 0 from the function [latex]f[\/latex], we need to decide whether a plus or a minus sign will work to satisfy [latex]g\\left(2\\right)=f\\left(x - 2\\right)=f\\left(0\\right)=0[\/latex]. For this to work, we will need to <em>subtract<\/em> 2 units from our input values.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Write a formula for the graph shown in Figure 24, which is a transformation of the toolkit square root function.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010518\/CNX_Precalc_Figure_01_05_0112.jpg\" alt=\"Graph of a square root function transposed right one unit and up 2.\" width=\"487\" height=\"292\" \/><figcaption class=\"wp-caption-text\"><b>Figure 24<\/b><\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q745825\">Show Solution<\/button><\/p>\n<div id=\"q745825\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137549429\">The graph of the toolkit function starts at the origin, so this graph has been shifted 1 to the right and up 2. In function notation, we could write that as<\/p>\n<p style=\"text-align: center;\">[latex]h\\left(x\\right)=f\\left(x - 1\\right)+2[\/latex]<\/p>\n<p id=\"fs-id1165135175226\">Using the formula for the square root function, we can write<\/p>\n<p style=\"text-align: center;\">[latex]h\\left(x\\right)=\\sqrt{x - 1}+2[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135500702\">Note that this transformation has changed the domain and range of the function. This new graph has domain [latex]\\left[1,\\infty \\right)[\/latex] and range [latex]\\left[2,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h3>Writing Functions Given the Transformed Graph<\/h3>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<p id=\"fs-id1165137643555\"><iframe id=\"ohm317561\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=317561&theme=lumen&iframe_resize_id=ohm317561&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe id=\"ohm33047\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=33047&theme=lumen&iframe_resize_id=ohm33047&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe id=\"ohm317562\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=317562&theme=lumen&iframe_resize_id=ohm317562&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<p id=\"fs-id1165137653909\">Write a formula for a transformation of the toolkit reciprocal function [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] that shifts the function\u2019s graph one unit to the right and one unit up.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44868\">Show Solution<\/button><\/p>\n<div id=\"q44868\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]g\\left(x\\right)=\\frac{1}{x - 1}+1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div><button><span aria-hidden=\"true\">&times;<\/span><span class=\"screen-reader-text\">Close definition<\/span><\/button><\/div><\/template><\/div>","protected":false},"author":67,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":74,"module-header":"background_you_need","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2556"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/67"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2556\/revisions"}],"predecessor-version":[{"id":5290,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2556\/revisions\/5290"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/74"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2556\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=2556"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=2556"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=2556"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=2556"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}