help please I i

Question

A family of parabolas satisfies the rule y = ax² + 2x + a.

1. Express ax²+2x+a in the form a(x+b)²+c for real numbers b and c.
2. Give the coordinates of the turning point of the graph of y=ax²+2x+a in terms of a.
3. For which values of a is ax²+2x+a a perfect square?
4. For which values of a does the graph of y=ax²+2x+a have two x-axis intercepts?

Answer 1

ax^2 +2x + a

= a(x^2 +2x/a) + a

= a(x^2 +2x/a + 1/a^2) + a - 1/a

= a(x+ 1/a)^2 + (a^2-1)/a

b= 1/a,

c= (a^2-a)/a = a-1/a = a-b

c=a-b

turning point = vertex = max point = (-1/a, a-1/a)

or take the derivarive, set = 0

y'= 2ax+2=0

ax=-1

x=-1/a

y=a(-1/a)^2 +2(-1/a)+a= 1/a -2/a+a= a-1/a

(x,y)=(-1/a, a-1/a)

when a=1, x^2+2x+1 =(x+1)^2 = a perfect square

two x-intercepts whenever the discriminant >0

when 4-4a^2>0

when a^2<1

when -1<a<1

when |a|<1

if |a|=1 there is one x-intercept

if |a|>1 there are no real x intercepts