// Numbas version: exam_results_page_options {"name": "Jamie's copy of Graphing exponentials of the form y=b^x with b>1", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {"graph1": {"type": "html", "parameters": [["quad", "number"]], "definition": "var div = Numbas.extensions.jsxgraph.makeBoard('300px','300px',{boundingBox:[-12,12,12,-12],grid:true,axis:false});\nvar board = div.board;\n\n// create the x-axis.\nvar xaxis = board.create('line',[[0,0],[1,0]], { strokeColor: 'black', fixed: true});\nvar xticks = board.create('ticks',[xaxis,2],{\n drawLabels: true,\n label: {offset: [-4, -10]},\n minorTicks: 0\n});\n\n// create the y-axis\nvar yaxis = board.create('line',[[0,0],[0,1]], { strokeColor: 'black', fixed: true });\nvar yticks = board.create('ticks',[yaxis,2],{\ndrawLabels: true,\nlabel: {offset: [-20, 0]},\nminorTicks: 0\n});\n\nb = Numbas.jme.unwrapValue(scope.variables.b);\n\n\n\nif(quad==1){board.create('functiongraph',[function(x){ return Math.pow(b,x)}],{strokeWidth:2});}\nif(quad==2){board.create('functiongraph',[function(x){ return Math.pow(1/b,x)}],{strokeWidth:2});}\nif(quad==3){board.create('functiongraph',[function(x){ return -Math.pow(1/b,x)}],{strokeWidth:2});}\nif(quad==4){board.create('functiongraph',[function(x){ return -Math.pow(b,x)}],{strokeWidth:2});}\n\nreturn div;", "language": "javascript"}}, "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "

The easiest type of exponential to graph where the base is greater than 1 and no transformations take place.

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The following questions will gauge your understanding of exponentials and how to graph them. 

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The exponential you will be working with for this question is \\[y=\\var{b}^x.\\]

\n

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a) To find the $y$-intercept, substitute $x=0$ into the equation: $y=\\var{b}^0=1$. Therefore, the $y$-intercept is the point $(0,1)$.

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b) Substitute $x=1$ into the equation: $y=\\var{b}^1=\\var{b}$. Therefore, another easily found point is $(1,\\var{b})$.

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c) Let's investigate what happens to the value of $y$ when we add 1 to the value of $x$:

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\\[\\var{b}^{x+1}=\\var{b}^x\\var{b}^1=\\var{b}^x\\var{b}\\]  That is, the old $y$ value is multiplied by $\\var{b}$, so we can say that $y$ is increased by a factor of $\\var{b}$.

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d) Since $y=\\var{b}^x$ is an exponential and as $x$ increases $y$ increases without bound, we call this exponential growth.

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e) An asymptote is a line or curve that approaches a given curve arbitrarily closely. For the curve $y=\\var{b}^x$ the smaller $x$ gets, the closer $y$ gets to $0$. In other words as $x$ approaches negative infinity, $y$ approaches $0$. This means that the asymptote for $y=\\var{b}^x$ is the line $y=0$ (the $x$-axis).

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f) Given all the information above, it should be clear that the graph should look like 

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{graph1(1)}

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The $y$-intercept of $y=\\var{b}^x$ is the point $\\large($0,[[0]]$\\large)$.

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Another easily found point on the curve is ${\\large(}1,$ [[0]]$\\large)$.

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increases by 1.

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decreases by 1.

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increases by {b-1}.

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decreases by {b-1}.

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increases by a factor of {b}.

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decreases by a factor of {b}.

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Given $y=\\var{b}^x$, everytime $x$ increases by 1, $y$  [[0]].

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exponential growth

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exponential decay

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Would $y=\\var{b}^x$ best be described as exponential decay or exponential growth?

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{graph1(1)}

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{graph1(2)}

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{graph1(3)}

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{graph1(4)}

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Which graph best represents $y=\\var{b}^x$?

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