Question

Find the missing coefficients


y=ax^2-10x+c
vertex :(-5,20)

Answer 1

remember, for
`y=ax^2+bx+c` the x value of the vertex is
`(-b)/(2a)`

so
given
x value of vertex is -5
and

`y=ax^2-10x+c`

`(-b)/(2a)=(-(-10))/(2a)=(10)/(2a)=`

`(5)/(a)=-5`
multiply both sides by a
5=-5a
divvide both sides by -5
-1=a
a=-1



`y=-1x^2-10x+c`
comlete the square
remember if we do

`y=a(x-h)^2+k` (h,k) is the vertex
I know, we can subsitute the known values of te vertex
-5 for h and 20 for k then expand


`y=-1(x-(-5))^2+20`

`y=-1(x+5)^2+20`

`y=-1(x^2+10x+25)+20`

`y=-1x^2-10x-25+20`

`y=-1x^2-10x-5`

a=-1
c=-5