What is the equation of the parabola that has a vertex at # (-3, 6) # and passes through point # (1,9) #?

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Answer 1 · 2015-12-19T16:39:21

#f(x) = 3/16x^2+ 9/8x + 123/16#

The parabola #f# is written as #ax^2 + bx + c# such that #a != 0#.

1st of all, we know this parabol has a vertex at #x=-3# so #f'(-3) = 0#. It already gives us #b# in function of #a#.

#f'(x) = 2ax + b# so #f'(-3) = 0 iff -6a + b = 0 iff b = 6a#

We now have to deal with two unknown parameters, #a# and #c#. In order to find them, we need to solve the following linear system :

#6 = 9a - 18a + c; 9 = a + 6a + c iff 6 = -9a + c;9 = 7a + c#

We now substract the 1st line to the 2nd one in the 2nd line :

#6 = -9a + c;3 = 16a# so we now know that #a = 3/16#.

We replace #a# by its value in the 1st equation :

#6 = -9a + c iff c = 6 + 9*(3/16) iff c = 123/16# and #b = 6a iff b = 9/8#.