How do you find a function whose graph is a parabola with vertex \[( - 2,2)\] and that passes through the point $ (1, - 4) $ ?

Hint: Parabola: It is one of the “Conic sections” in which a special curve, shaped like an arch. Any point on a parabola is at an equal distance form….
A fixed point (the focus) and a fixed straight line (the directrix).
The equation of a parabola in vertex form is,
  $ y = a{(x - h)^2} + k $
Where $ (h,k) $ are the coordination of the vertex and $ a, $ is a constant.

Complete step by step solution:
The equation of a parabola in vertex form is
  $ y = a{(x - h)^2} + k $
As given in question $ (h,k) = ( - 2,2) $ .
By keeping this value in the above mentioned equation
  $ \Rightarrow y = a{(x + 2)^2} + 2 $
To find $ a $ , substitute $ x = 1,y = - 4 $ from $ (1, - 4) $ into the partial equation.
  $ a{(1 + 2)^2} + 2 = - 4 $
  $ \Rightarrow 9a = - 6 $
  $ \Rightarrow a = - \dfrac{2}{3} $
Keeping this value in the partial equation.
  $ \Rightarrow y = - \dfrac{2}{3}(x + 2) + 2 $
Hence, the function whose graph is a parabola with vertex \[( - 2,2)\] and that passes through the point $ (1, - 4) $ is $ y = - \dfrac{2}{3}(x + 2) + 2 $ .
So, the correct answer is “ $ y = - \dfrac{2}{3}(x + 2) + 2 $ ”.

Note: parabola have the property that, if they are made of material that reflects light, then light that travel parallel to the axis of symmetry of a parabola and strikes its concave side is reflected to its focus, regardless occur, conversely, light that originates from a point source at the focus is reflected into a parallel beam, leaving the parabola parallel to the axis of symmetry. The same efforts occur with sound and other waves. This reflective property is the basis of many practical uses of parabolas.