How do you write a quadratic function in vertex form whose has vertex (-1/2, 2/3) and passes through point (3/2, 14/3)?

Question

Kennedy Adams · Accepted Answer

#y=(x+1/2)^2+2/3# #"the equation of a parabola in "color(blue)"vertex form"# is. #color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))# #"where "(h,k)" are the coordinates of the vertex and a"# #"is a multiplier"# #"here "(h,k)=(-1/2,2/3)# #rArry=a(x+1/2)^2+2/3# #"to find a substitute "(3/2,14/3)" into the equation"# #14/3=4a+2/3rArr4a=4rArra=1# #rArry=(x+1/2)^2+2/3larrcolor(blue)"in vertex form"#

Savannah Anderson · Answer

undefined To write a quadratic function in vertex form when given the vertex and a point that it passes through, you can use the vertex form equation: $ f(x) = a(x - h)^2 + k $, where $ (h, k) $ is the vertex.

Given the vertex $ (-1/2, 2/3) $, we have $ h = -1/2 $ and $ k = 2/3 $.

Substituting the vertex values into the vertex form equation, we get: $ f(x) = a(x + 1/2)^2 + 2/3 $.

Now, we need to find the value of $ a $.

Using the point $ (3/2, 14/3) $, substitute $ x = 3/2 $ and $ f(x) = 14/3 $ into the equation: $ 14/3 = a(3/2 + 1/2)^2 + 2/3 $.

Solve for $ a $:

$ 14/3 = a(2)^2 + 2/3 $
$ 14/3 = 4a + 2/3 $
$ 14/3 - 2/3 = 4a $
$ 4 = 4a $
$ a = 1 $

Therefore, the quadratic function in vertex form is: $ f(x) = (x + 1/2)^2 + 2/3 $.