Exam 14: Review

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Find the general form of the equation of the the circle. -Center at the point (−4,−3)( - 4 , - 3 ) ; containing the point (−3,3)( - 3,3 )

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Determine whether the graph of the equation is symmetric with respect to the x-axis, the y-axis, and/or the origin. - y2−x−1=0y ^ { 2 } - x - 1 = 0

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Determine whether the given point is on the graph of the equation. -Equation: x2−y2=16x ^ { 2 } - y ^ { 2 } = 16 Point: (22,22)( 2 \sqrt { 2 } , 2 \sqrt { 2 } )

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Find the general form of the equation of the the circle. -Center at the point (2,−3)( 2 , - 3 ) ; containing the point (5,−3)( 5 , - 3 )

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Find the slope of the line. -Find the slope of the line. -

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List the intercepts of the graph.Tell whether the graph is symmetric with respect to the x-axis, y-axis, origin, or none of these. - List the intercepts of the graph.Tell whether the graph is symmetric with respect to the x-axis, y-axis, origin, or none of these. -  A) intercept:  ( 0,9 )  symmetric with respect to  x -axis  B) intercept:  ( 9,0 )  no symmetry  C) intercept:  ( 0,9 )  no symmetry  D) intercept:  ( 9,0 )  symmetric with respect to  \mathrm { y } -axis A) intercept: (0,9)( 0,9 ) symmetric with respect to xx -axis B) intercept: (9,0)( 9,0 ) no symmetry C) intercept: (0,9)( 0,9 ) no symmetry D) intercept: (9,0)( 9,0 ) symmetric with respect to y\mathrm { y } -axis

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Graph the circle with radius r and center (h, k). - r=6;(h,k)=(0,4)r=6 ;(h, k)=(0,4)  Graph the circle with radius r and center (h, k). - r=6 ;(h, k)=(0,4)        A)   B)    C)   D)      A)  Graph the circle with radius r and center (h, k). - r=6 ;(h, k)=(0,4)        A)   B)    C)   D)      B)  Graph the circle with radius r and center (h, k). - r=6 ;(h, k)=(0,4)        A)   B)    C)   D)      C)  Graph the circle with radius r and center (h, k). - r=6 ;(h, k)=(0,4)        A)   B)    C)   D)      D)  Graph the circle with radius r and center (h, k). - r=6 ;(h, k)=(0,4)        A)   B)    C)   D)

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Find the equation of the line in slope-intercept form. -Find the equation of the line in slope-intercept form. -

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Graph the equation by plotting points. - y=xy=\sqrt{x}  Graph the equation by plotting points. - y=\sqrt{x}      A)   B)   C)   D)    A)  Graph the equation by plotting points. - y=\sqrt{x}      A)   B)   C)   D)    B)  Graph the equation by plotting points. - y=\sqrt{x}      A)   B)   C)   D)    C)  Graph the equation by plotting points. - y=\sqrt{x}      A)   B)   C)   D)    D)  Graph the equation by plotting points. - y=\sqrt{x}      A)   B)   C)   D)

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Plot the point. - (−3,−1)( - 3 , - 1 )  Plot the point. - ( - 3 , - 1 )      A)   B)    C)   D)    A)  Plot the point. - ( - 3 , - 1 )      A)   B)    C)   D)    B)  Plot the point. - ( - 3 , - 1 )      A)   B)    C)   D)    C)  Plot the point. - ( - 3 , - 1 )      A)   B)    C)   D)    D)  Plot the point. - ( - 3 , - 1 )      A)   B)    C)   D)

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Find the slope of the line containing the two points. -(-4, -3); (-4, -2)

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Write the standard form of the equation of the circle with radius r and center (h, k). - r=19;(h,k)=(7,−10)\mathrm { r } = \sqrt { 19 } ; ( \mathrm { h } , \mathrm { k } ) = ( 7 , - 10 )

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Solve the problem. -Earth is represented on a map of the solar system so that its surface is a circle with the equation x2+y2+8x+4y−3,824=0x ^ { 2 } + y ^ { 2 } + 8 x + 4 y - 3,824 = 0 . A weather satellite circles 0.40.4 units above the Earth with the center of its circular orbit at the center of the Earth. Find the general form of the equation for the orbit of the satellite on this map.

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Find an equation for the line with the given properties. -Perpendicular to the line -3x - 4y = 15; containing the point (3, 2)

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Find an equation for the line with the given properties. -Perpendicular to the line -2x - 5y = -3; y-intercept = 5

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Find the center (h, k) and radius r of the circle with the given equation. - x2+y2=9x ^ { 2 } + y ^ { 2 } = 9

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List the intercepts of the graph. - List the intercepts of the graph. -  A)  \left( 0 , - \frac { \pi } { 2 } \right) , ( 0 , - 2 ) , \left( 0 , \frac { \pi } { 2 } \right)  B)  \left( - \frac { \pi } { 2 } , 0 \right) , ( - 2,0 ) , \left( \frac { \pi } { 2 } , 0 \right)  C)  \left( 0 , - \frac { \pi } { 2 } \right) , ( - 2,0 ) , \left( 0 , \frac { \pi } { 2 } \right)  D)  \left( - \frac { \pi } { 2 } , 0 \right) , ( 0 , - 2 ) , \left( \frac { \pi } { 2 } , 0 \right) A) (0,−π2),(0,−2),(0,π2)\left( 0 , - \frac { \pi } { 2 } \right) , ( 0 , - 2 ) , \left( 0 , \frac { \pi } { 2 } \right) B) (−π2,0),(−2,0),(π2,0)\left( - \frac { \pi } { 2 } , 0 \right) , ( - 2,0 ) , \left( \frac { \pi } { 2 } , 0 \right) C) (0,−π2),(−2,0),(0,π2)\left( 0 , - \frac { \pi } { 2 } \right) , ( - 2,0 ) , \left( 0 , \frac { \pi } { 2 } \right) D) (−π2,0),(0,−2),(π2,0)\left( - \frac { \pi } { 2 } , 0 \right) , ( 0 , - 2 ) , \left( \frac { \pi } { 2 } , 0 \right)

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Solve the problem. -Find all values of k so that the given points are 29\sqrt { 29 } units apart. (-5, 5), (k, 0)

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Find an equation for the line with the given properties. -Perpendicular to the line x - 2y = 2; containing the point (5, 3) A) y=2x−13y = 2 x - 13 B) y=−2x+13y = - 2 x + 13 C) y=−12x−132y = - \frac { 1 } { 2 } x - \frac { 13 } { 2 } D) y=−2x−13y = - 2 x - 13

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Solve the problem. -A middle school's baseball playing field is a square, 55 feet on a side. How far is it directly from home plate to second base (the diagonal of the square)? If necessary, round to the nearest foot.

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