f(x)=x^2=10x+35 what is the value of h when the function is converted to vertex form
I hope this helps you
y=x^2-10x+35
y=x^2-10x+25+10
y-10=(x-5)^2
y=(x-5)^2+10
h= -(-5)=5
k =10
vertex form is
f (x) = a(x - h)2 + k, where (h, k) is the vertex of the parabola
f(x)=x^2+10x+35
x^2+10x+25-25+35
x^2+5x+5x+25+10
x(x+5)5(x+5)+10
(x+5)^2+10
where h=-5