Question

Use the given vertex ( 5/2, -3/4) and point ( -2, 4) to write the equation of the quadratic function. How Do I Solve This Problem?

Answer 1

Equations are `361x=-72y^2-108y+862` or `81y=19x^2-95x+58`

Explanation:

Equation of quadratic function with `(h,k)` as vertex is

`y=a(x-h)^2+k` or `x=b(y-k)^2+h`

Let us consider first case . As vertex is `(5/2,-3/4)`, the equation is of type `y=a(x-5/2)^2-3/4`. As it passes through `(-2,4)`, we have

`4=a(-2-5/2)^2-3/4` i.e. `4=axx81/4-3/4`

or `81/4a=4+3/4=19/4` i.e. `a=19/81`

and equation is `y=19/81(x-5/2)^2-3/4`

or `81y=19(x^2-5x+25/4)-243/4=19x^2-95x+58`

Second Case - As vertex is `(5/2,-3/4)`, the equation is of type `x=b(y+3/4)^2+5/2`. As it passes through `(-2,4)`, we have

`-2=b(4+3/4)^2+5/2` or `361/16b=-2-5/2=-9/2`

or `b=-9/2xx16/361=-72/361`

and equation is `x=-72/361(y+3/4)^2+5/2`

or `361x=-72(y^2+3/2y+9/16)+1805/2`

or `361x=-72y^2-108y-81/2+1805/2`

or `361x=-72y^2-108y+862`

graph{(72y^2+108y-862+361x)(19x^2-95x+58-81y)=0 [-9.27, 10.73, -4.74, 5.26]}