SOLUTION: Determine the quadratic function, f(x) = ax^2 + bx + c, whose vertex is (1, 8) and passes through the point (-3, 4)

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Question 540835: Determine the quadratic function, f(x) = ax^2 + bx + c, whose vertex is (1, 8) and passes through the point (-3, 4)
Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
Determine the quadratic function, f(x) = ax^2 + bx + c, whose vertex is (1, 8) and passes through the point (-3, 4)
It has the form

f(x) = a(x - h)² + k where the vertex is (h,k),

Since we are given that the vertex is (1,8), we have

f(x) = a(x - 1)² + 8

Since the parabola passes through the point (x,y) = (-3,4), we will
substitute -3 for x and since y = f(x) we will substitute 4 for f(x):

4 = a(-3 - 1)² + 8

4 = a(-4)² + 8

4 = a(16) + 8

4 = 16a + 8

-4 = 16a

 = a

 = a

We substitute  for a in

f(x) = a(x - 1)² + 8

f(x) = (x - 1)² + 8

That is in vertex form.  To get it in standard form  f(x) = ax² + bx + c

f(x) = (x - 1)(x - 1) + 8

f(x) = (x² - 2x + 1) + 8

f(x) = x² - ·2x + ·1 + 8

f(x) = x² + x -  + 8

f(x) = x² + x -  + 32/4

f(x) = x² + x + 




Edwin




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