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A parabola has one xintercept at (– 4, 0). If the vertex is at (2, 5
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15 Mar 2015, 22:09
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Re: A parabola has one xintercept at (– 4, 0). If the vertex is at (2, 5
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16 Mar 2015, 02:11
Answer is D (8) Parabola y= ax^2+bx+c x intercept at (4,0) then 16a4b+c=0 <=> c =4b16a (1) Vertex (b/2a;c b^2/4a)=(2,5) then b/2a=2 <=> b= 4a (2) (1)+(2)=> c=8b (3) c b^2/4a = 5 (4) From (2)(3)(4) we have 8b b^2/(b)=5 <=> 9b= 5 then b =5/9 then c =40/9 and a = 5/36 the second intercept x we have 5/36 x^2+ 5/9x+40/9=0 <=> x^24x32=0 Using back solving,we have x =8; 8^24*832=0
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Re: A parabola has one xintercept at (– 4, 0). If the vertex is at (2, 5
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16 Mar 2015, 09:05
Parabola is symmetric about a line parallel to Y axis in this case. Therefore the X coordinate distance of the other intercept is same from the vertex as the given intercept. The distance = 2(4)=6 , There the other coordinate is 2+6 = 8.



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Re: A parabola has one xintercept at (– 4, 0). If the vertex is at (2, 5
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16 Mar 2015, 16:28
awanish1r wrote: Parabola is symmetric about a line parallel to Y axis in this case. Therefore the X coordinate distance of the other intercept is same from the vertex as the given intercept. The distance = 2(4)=6 , There the other coordinate is 2+6 = 8. u are right. my way is silly too many years not study math sense is lost
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Re: A parabola has one xintercept at (– 4, 0). If the vertex is at (2, 5
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23 Mar 2015, 02:50
Bunuel wrote: A parabola has one xintercept at (– 4, 0). If the vertex is at (2, 5), find the other xintercept.
A. 5 B. 6 C. 7 D. 8 E. 9
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:The line of symmetry of a parabola always passes through the vertex, so the equation of the line of symmetry is the vertical line x = 2. The two xintercepts are symmetrical around this line. The point (– 4, 0) is six units to the left of the symmetry line, so the other should be six units to right, at (8, 0). Attachment:
ppocg_img12.png [ 15.61 KiB  Viewed 6774 times ]
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Re: A parabola has one xintercept at (– 4, 0). If the vertex is at (2, 5
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14 Jul 2016, 05:22
Bunuel wrote: A parabola has one xintercept at (– 4, 0). If the vertex is at (2, 5), find the other xintercept.
A. 5 B. 6 C. 7 D. 8 E. 9
Kudos for a correct solution. The coordinates makes it clear that we are dealing with a parabola like a "inverted U" or a "straight A" One arm of the parabola is cutting the x axis at 4, the vortex is 2 the distance between the arm and vortex is 6 units . THE OTHER ARM WILL BE A MIRROR IMAGE OF THE FIRST ARM... AN IMAGINARY VERTICAL LINE DRAWN THROUGH THE VORTEX WILL ACTS AS THE AXIS OF SYMMETRY. The vortex is at x=2. Now the other arm will be 6 unit away from x=2 There fore the other arm will be at X=2+6 = 8 The x coordinate of the other arm will be 8 ANSWER IS D
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Re: A parabola has one xintercept at (– 4, 0). If the vertex is at (2, 5
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26 Jul 2017, 22:16
The point of vertex is the midpoint for the parabola and hence we can consider the midpoint formula to calculate the other interscept. Am i correct in my concept . Kindly help.



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Re: A parabola has one xintercept at (– 4, 0). If the vertex is at (2, 5
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26 Jul 2017, 23:01
Bunuel wrote: A parabola has one xintercept at (– 4, 0). If the vertex is at (2, 5), find the other xintercept.
A. 5 B. 6 C. 7 D. 8 E. 9
Kudos for a correct solution. Vertex of the parabola is (2,5) . The x coordinate is 2. So parabola is symmetrical about x =2 Parabola has x  intercept as (4,0 ). Sodistance between x =2 and xinctercept of arabola = 2(4) = 6 other x  intercept = (2+6,0) = (8.0) Answer D
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Re: A parabola has one xintercept at (– 4, 0). If the vertex is at (2, 5
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25 Nov 2018, 16:20
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Re: A parabola has one xintercept at (– 4, 0). If the vertex is at (2, 5 &nbs
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