Find the equation algebraically, of the parabola which passes through the points (10,-14) and (-2,10), and whose axis of symmetry is the equation x=2, using vertex form.
+118696 Find the equation algebraically, of the parabola which passes through the points (10,-14) and (-2,10), and whose axis of symmetry is the equation x=2, using vertex form.
Interesting question
(10,-14), (1)
(-2,10) (2)
Vertex(2, k) (3)
(x−h)2=4a(y−k) where (h,k) is the vertex
(x−2)2=4a(y−k)
Using (10,-14) we have 64=4a(−14−k)
Using (-2,10) we have 16=4a(10−k)
6416=4a(−14−k)4a(10−k)4=−14−k10−k4=−14−k10−k40−4k=−14−k54=3kk=18
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(x−2)2=4a(y−18)
(10,−14)64=4a(−32)→−2=4a→a=−0.5check(−2,10)16=4a(−8)→a=−0.5
So the equation is
(x−2)2=−2(y−18)
And that is that. Can I have a thumbs up now please. OR if you don't understand ask for clarification. 
+118696 Find the equation algebraically, of the parabola which passes through the points (10,-14) and (-2,10), and whose axis of symmetry is the equation x=2, using vertex form.
Interesting question
(10,-14), (1)
(-2,10) (2)
Vertex(2, k) (3)
(x−h)2=4a(y−k) where (h,k) is the vertex
(x−2)2=4a(y−k)
Using (10,-14) we have 64=4a(−14−k)
Using (-2,10) we have 16=4a(10−k)
6416=4a(−14−k)4a(10−k)4=−14−k10−k4=−14−k10−k40−4k=−14−k54=3kk=18
---------------------
(x−2)2=4a(y−18)
(10,−14)64=4a(−32)→−2=4a→a=−0.5check(−2,10)16=4a(−8)→a=−0.5
So the equation is
(x−2)2=−2(y−18)
And that is that. Can I have a thumbs up now please. OR if you don't understand ask for clarification.
