SOLUTION: Find the equation of the quadratic function whose graph passes through the point (-1,3) and has vertex (2,-4) . ?

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Question 725980: Find the equation of the quadratic function whose graph passes through the point (-1,3) and has vertex (2,-4) .
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Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
If equation has the form +y+=+a%2Ax%5E2+%2B+b%2Ax+%2B+c+,
then the x co-ordinate of the vertex is at
+x%5Bv%5D+=+-b%2F%28a%29+
given: vertex at (2,-4)
+-b%2F%282a%29+=+2+
+-b+=+4a+
+b+=+-4a+
also:
+-4+=+a%2Ax%5E2+-+4a%2Ax+%2B+c+
+-4+=+a%2A2%5E2+-+4a%2A2+%2B+c+
+-4+=+4a+-+8a+%2B+c%0D%0A%7B%7B%7B+-4a+%2B+c+=+-4+
(1) +4a+-+c+=+4+
-----------------------
Also given: (-1,3) is a solution
+y+=+a%2Ax%5E2+%2B+b%2Ax+%2B+c+
+3+=+a%2A%28-1%29%5E2+-+4a%2A%28-1%29+%2B+c+
+3+=+a+%2B+4a+%2B+c+
(2) +5a+%2B+c+=+3+
-----------------
Add (1) and (2)
(1) +4a+-+c+=+4+
(2) +5a+%2B+c+=+3+
+9a+=+7+
+a+=+7%2F9+
and, since
+b+=+-4a+
+b+=+-4%2A%287%2F9%29+
+b+=+-28%2F9+
--------------
also
(2) +5a+%2B+c+=+3+
(2) +35%2F9+%2B+c+=+3+
(2) +c+=+27%2F9+-+35%2F9+
(2) +c+=+-8%2F9+
The equation is:
+y+=+%287%2F9%29%2Ax%5E2+-+%2828%2F9%29%2Ax+-+8%2F9+
--------------------------------
check: is the vertex at (2,-4) ?
+-b%2F%282a%29+=+2+
+%2828%2F9%29+%2F+%282%2A%287%2F9%29%29+=+2+
+%2828%2F9%29%2A%289%2F14%29+=+2+
+2+=+2+
+-4+=+%287%2F9%29%2A4+-+%2828%2F9%29%2A2+-+8%2F9+
+-4+=+28%2F9+-+56%2F9+-+8%2F9+
+-36+=+28+-56+-8+
+-36+=+-36+
OK
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Does it go through (-1,3) ?
+y+=+%287%2F9%29%2Ax%5E2+-+%2828%2F9%29%2Ax+-+8%2F9+
+3+=+%287%2F9%29%2A%28-1%29%5E2+-+%2828%2F9%29%2A%28-1%29+-+8%2F9+
+3+=+7%2F9+%2B+28%2F9+-+8%2F9+
+27+=+7+%2B+28+-+8+
+27+=+27+
OK