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 $\frac{-b}{2a}$

so
given
x value of vertex is -5
and
$y=ax^2-10x+c$
$\frac{-b}{2a}=\frac{-(-10)}{2a}=\frac{10}{2a}=$
$\frac{5}{a}=-5$
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