Exam 3: Graphs and Functions

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Describe how the graph of the equation relates to the graph of y y=x3y = \sqrt [ 3 ] { x } . - f(x)=(2x)2f(x)=(2 x)^{2}  Describe how the graph of the equation relates to the graph of y  y = \sqrt [ 3 ] { x }  . - f(x)=(2 x)^{2}

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The line graph shows the recorded hourly temperatures in degrees Fahrenheit at an airport. The line graph shows the recorded hourly temperatures in degrees Fahrenheit at an airport.   -At what time was the temperature the highest? -At what time was the temperature the highest?

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Graph the line described. -through (2,9)( - 2,9 ) ; undefined slope  Graph the line described. -through  ( - 2,9 ) ; undefined slope

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Give the domain and range of the relation. -Give the domain and range of the relation. -

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Describe how the graph of the equation relates to the graph of y y=x3y = \sqrt [ 3 ] { x } . - y=5x21y=5 x^{2}-1  Describe how the graph of the equation relates to the graph of y  y = \sqrt [ 3 ] { x }  . - y=5 x^{2}-1

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Give the domain and range of the relation. -Find f(4)f ( 4 ) when f(x)=5x2+2x2f ( x ) = 5 x ^ { 2 } + 2 x - 2

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Give the domain and range of the relation. -Annual New Telemarketing Companies Year Number 1993 52 1994 102 1995 187 1996 170 1997 218

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A circle has a diameter with endpoints (2,1)( - 2,1 ) and (22,11)( 22,11 ) . Find the radius.

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Find the requested value. - f(8)f ( 8 ) for f(x)={6x+1, if x<18x, if 8x1184x, if x>11f ( x ) = \left\{ \begin{array} { l l } 6 x + 1 , & \text { if } x < 1 \\ 8 x , & \text { if } 8 \leq x \leq 11 \\ 8 - 4 x , & \text { if } x > 11 \end{array} \right.

(Multiple Choice)
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Graph the point symmetric to the given point. -  Plot the point (5,0), then plot the point that is symmetric to (5,0) with respect to the x-axis. \text { Plot the point } ( - 5,0 ) \text {, then plot the point that is symmetric to } ( - 5,0 ) \text { with respect to the } x \text {-axis. }  Graph the point symmetric to the given point. - \text { Plot the point } ( - 5,0 ) \text {, then plot the point that is symmetric to } ( - 5,0 ) \text { with respect to the } x \text {-axis. }

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Determine whether the three points are the vertices of a right triangle. - (1,1),(5,1),(4,4)( - 1 , - 1 ) , ( 5,1 ) , ( 4 , - 4 )

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The graphs of functions f\mathrm { f } and g\mathrm { g } are shown. Use these graphs to find (fg)(1)( \mathrm { fg } ) ( 1 ) .  The graphs of functions  \mathrm { f }  and  \mathrm { g }  are shown. Use these graphs to find  ( \mathrm { fg } ) ( 1 ) .     y = f ( x )   y = g ( x ) y=f(x)y = f ( x ) y=g(x)y = g ( x )

(Multiple Choice)
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Graph the line and give the domain and the range. - x+4=0x+4=0  Graph the line and give the domain and the range. - x+4=0

(Multiple Choice)
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Find the requested function value. -Find (gf)(5)( g \circ f ) ( - 5 ) when f(x)=9x5f ( x ) = - 9 x - 5 and g(x)=9x29x1g ( x ) = 9 x ^ { 2 } - 9 x - 1 .

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Describe the transformations and give the equation for the graph. -Find (fg)(2)( f g ) ( 2 ) when f(x)=x6f ( x ) = x - 6 and g(x)=3x2+11x7g ( x ) = - 3 x ^ { 2 } + 11 x - 7

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Decide whether or not the equation has a circle as its graph. If it does not, describe the graph. - x2+y2+16x+12y+211=0x ^ { 2 } + y ^ { 2 } + 16 x + 12 y + 211 = 0

(Multiple Choice)
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Graph the point symmetric to the given point. -  Plot the point (0,1), then plot the point that is symmetric to (0,1) with respect to the y-axis. \text { Plot the point } ( 0 , - 1 ) \text {, then plot the point that is symmetric to } ( 0 , - 1 ) \text { with respect to the } y \text {-axis. }  Graph the point symmetric to the given point. - \text { Plot the point } ( 0 , - 1 ) \text {, then plot the point that is symmetric to } ( 0 , - 1 ) \text { with respect to the } y \text {-axis. }

(Multiple Choice)
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Determine the intervals of the domain over which the function is continuous. -Determine the intervals of the domain over which the function is continuous. -

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Find the specified domain. -Find the domain of (f+g)(x)( f + g ) ( x ) when f(x)=8x2f ( x ) = \sqrt { 8 x - 2 } and g(x)=1xg ( x ) = \frac { 1 } { x }

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Consider the function h as defined. Find functions f and g so tha (fg)(x)=h(x)( f \circ g ) ( x ) = h ( x ) - h(x)=2x+2h ( x ) = | 2 x + 2 |

(Multiple Choice)
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