How do you find two quadratic function one that opens up and one that opens downward whose graphs have intercepts (4,0), (8,0)?
1 Answer
Explanation:
There are an infinite number of quadratic functions possible.
Since we know the x-intercepts (4 ,0) and (8 ,0)
rArr"roots are " x=4" and " x=8⇒roots are x=4 and x=8
"and the factors are "(x-4),(x-8)and the factors are (x−4),(x−8) Hence the general equation is.
y=a(x-4)(x-8)y=a(x−4)(x−8) • If a > 0 , then graph opens up
• If a < 0 , then graph opens down
To find the particular function we require to know another point
For this example, lets choose a = 1 and a = - 1
a=1toy=x^2-12x+32larr" distributing brackets"a=1→y=x2−12x+32← distributing brackets
a=-1toy=-(x^2-12x+32)=-x^2+12x-32a=−1→y=−(x2−12x+32)=−x2+12x−32
graph{(y-x^2+12x-32)(y+x^2-12x+32)=0 [-10, 10, -5, 5]}
