How do you find the equation of the parabola whose vertex is at (1,-2), and that passes through the point (4,16)?

1 Answer
Alan P.
Sep 7, 2015

#y=2x^2-4x#
#color(white)("XXXXXXXX")#(see note below)

Explanation:

Assuming the parabola is a function in standard position
with vertex at #(1,-2)# and another point at #(4,16)#

The general vertex form is
#color(white)("XXX")y=m(x-a)^2+bcolor(white)("XXX")# with vertex at #(a,b)#

For the given case;
#color(white)("XXX")y=m(x-1)^2-2#
at #(4,16)#
#color(white)("XXX")16 = m(3)^2-2 = 9m-2#
#rarrcolor(white)("XXX")m=2#

Therefore the equation is
#color(white)("XXX")y=2(x-1)^2-2#
#color(white)("XXXX")= 2x^2-4x#

Note: the general vertex form assumes the parabola is in standard position (e.g. with a vertical axis of symmetry). Infinitely many alternate solutions are possible if this is not the case.
Three such parabolas are shown below:enter image source here

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
8248 views around the world
You can reuse this answer
Creative Commons License