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This is the second of our many quizzes on Graphing exponential functions. Click on this to access more such competitive quizzes on Algebra.

Graphing exponential functions Quiz 2

mathematics

 What will be the equation of the graph f?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e11b91a0-6ce7-452b-8577-5c4a03b0a43e2739659655595546349.png"/>

 Find y-intercept of the function and check for asymptotes using the graphs.

  Taking $f={{e}^{-2x}}$ $ \\ \Rightarrow f=\dfrac{1}{{{e}^{2x}}} \\ $ Putting $x=0$, we get $f=1$. $ \\ $ As we know that exponential function has horizontal asymptotes, $\\ $ So, $f\to 0$ as $x\to \infty $ and $f\to \infty $ as $x\to -\infty $ $\\ $ So, the function $f={{e}^{-2x}}$ will represent the given graph.

 What will be the equation of the given graph?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/81ce0c1a-f9c0-4bfe-9374-9e912c3932d13554624662309949884.png"/>

 Find y-intercept for the functions in the options by substituting $x=0$ and compare with the graph given to find the correct option.

  Let $f\left( x \right)={{3}^{x+1}}$ $\\ $  When $x=0$, $f\left( x \right)$ will be $\\ $ $\Rightarrow f\left( 0 \right)={{3}^{0+1}}={{3}^{1}}=3.$ $\\ $ Let  $g\left( x \right)={{3}^{x}}.$ $\\ $ When $x=0$, $g\left( x \right)$ will be $ \\ \Rightarrow g\left( 0 \right)={{3}^{0}}=1. \\ \Rightarrow g\left( 1 \right)={{3}^{1}}=3. \\ $ So, the graph will represent equation $f\left( x \right)={{3}^{x+1}}$,  $g\left( x \right)={{3}^{x}}.$

 What is the length of the dotted line?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/cf8b5ba8-ce23-4204-a1f9-8b3977fc709e3633358069844320504.png"/>

 Find the coordinates of the points given in the graph and use the distance between two points formula to get the required answer.

By the given graph we can locate the vertices of $A\left( -5,4 \right)$  and $B\left( 6,1 \right)$ $\\ $ Now, using the distance formula $\sqrt{{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}}$, $ \\ \Rightarrow AB=\sqrt{{{\left( 1-4 \right)}^{2}}+{{\left( 6-(-5) \right)}^{2}}} \\ \Rightarrow AB=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 11 \right)}^{2}}} \\ \Rightarrow AB=\sqrt{9+121}=\sqrt{130}$

 If given that $f(x) = 2^x \cdot 2^{-2x+2}, g(x) = 4^x \cdot 4^{-2x+2}$, then the point of intersection is 

 Take $f\left( x \right)=g\left( x \right)$ and find the value of $'x'$ which gives the abscissa of the point of intersection of both functions. Then substitute that value in the function to get the ordinate of the intersection point.

 Given, $f\left( x \right)={{2}^{x}}{{.2}^{-2x+2}}$ and $g\left( x \right)={{4}^{x}}{{.4}^{-2x+2}}$ $\\ $ $f\left( x \right)={{2}^{2-x}}$ and $g\left( x \right)={{4}^{2-x}}$………………. (i) $\\ $ For getting point of intersection $f\left( x \right)=g\left( x \right)$ $ \\ \Rightarrow {{2}^{2-x}}={{4}^{2-x}} \\ \Rightarrow {{2}^{2-x}}={{2}^{2(2-x)}} \\ \Rightarrow 2-x=2(2-x) \\ \Rightarrow 2-x=4-2x \\ \Rightarrow x=2 \\ $ Putting $x=2$ in both equations of (i) we get, $ \\ \Rightarrow f\left( x \right)={{2}^{2-2}}={{2}^{0}}=1 \\ \Rightarrow g\left( x \right)={{4}^{2-2}}={{4}^{0}}=1 \\ $ Therefore, the point of intersection is $\left( 1,1 \right)$.

 What is the distance between the points of intersection of $f(x) = 5^{2x^2} \cdot 5^{2x}, g(x) = 4^x$ and $h(x) = 3^{2x-6}, p(x) = 5^x - 5$ if $h(x)$ and $p(x)$ has the point of intersection on x-axis.

 Put $f\left( x \right)=g\left( x \right)$ to find the values of $'x'$ and use the distance formula between two points to get the required answer.

 Given, $f\left( x \right)={{5}^{2{{x}^{2}}}}{{.5}^{2x}}$ and $g\left( x \right)={{4}^{x}}$ $\\ $ For getting point of intersection $f\left( x \right)=g\left( x \right)$ $ \\ \Rightarrow {{5}^{2{{x}^{2}}}}{{.5}^{2x}}={{4}^{x}} \\ \Rightarrow {{5}^{2{{x}^{2}}+2x}}={{2}^{2}}^{x} \\  \Rightarrow 2{{x}^{2}}+2x(\log 5)=2x(\log 2) \\ \Rightarrow 2x\left( x+1 \right)\left( \log 5 \right)=2x\left( \log 2 \right) \\  \Rightarrow x\log 5+\log 5=\log 2 \\  \Rightarrow x\log 5=\log 2-\log 5 \\  \Rightarrow x=\dfrac{\log 2}{\log 5}-1 \\  \Rightarrow x=\dfrac{\text{0}\text{.301}}{0.699}-1 \\  \Rightarrow x=0.431-1 \\ \Rightarrow x=-0.569 \\ $ Putting $x=-0.569$ in $g\left( x \right)$ we get, $ \\ \Rightarrow g\left( -0.569 \right)={{4}^{-0.569}} \\ \Rightarrow g\left( -0.569 \right)=0.431 \\ $ The point of intersection of $f\left( x \right)$ and $g\left( x \right)$ is $~\left( -0.569,0.431 \right)$ $\\ $ Given, $h\left( x \right)={{3}^{2x+6}}$ and $p\left( x \right)={{5}^{x}}-5$ $\\ $ It has been said that the point of intersection is on $x-axis$, therefore $y=0$. $\\ $ So, we can conclude that $p\left( x \right)=0$ $ \\ \Rightarrow {{5}^{x}}-5={{0}^{{}}} \\ \Rightarrow {{5}^{x}}=5 \\ \Rightarrow x=1 \\ $ The point of intersection of $h\left( x \right)$ and $p\left( x \right)$ is $\left( 1,0 \right)$ $\\ $ Now, using the distance formula $\sqrt{\left( {{x}_{2}}-{{x}_{1}} \right)-\left( {{y}_{2}}-{{y}_{1}} \right)}$ , $ \\ = \sqrt{\left( 1-(-0.569) \right)-\left( 0-0.431 \right)}= \sqrt{2}$