How do you write a quadratic function in vertex form whose has vertex (3/4,2) and passes through point (2,41/8)?

2 Answers
Oct 4, 2017

#y=2(x-3/4)^2+2#

Explanation:

#"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)=(3/4,2)#

#rArry=a(x-3/4)^2+2#

#"substitute "(2,41/8)" into the equation for a"#

#rArr41/8=a(5/4)^2+2#

#rArr25/16a=25/8#

#rArra=25/8xx16/25=2#

#rArry=2(x-3/4)^2+2larrcolor(red)" in vertex form"#

Oct 4, 2017

The quadratic equation in vertex form is #y = 2(x-3/4)^2 + 2#

Explanation:

Vertex is at # (3/4,2) ;h=3/4, k=2#. The quadratic equation in

vertex form is #y=a(x-h^2)+k or y = a (x-3/4)^2 + 2# . The

point #(2,41/8)# satisfies the equation as it is on the parabola.

#:. 41/8 = a (2-3/4)^2 +2 or a (5/4)^2 =41/8 -2# or

#25/16*a = (41-16)/8 or cancel(25)/16*a = cancel(25)/8 :.#

# a=16/8=2 :.# Hence the quadratic equation in vertex form

is #y = 2(x-3/4)^2 + 2# [Ans]