Question

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

Answer 1

`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"`

Answer 2

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]