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
Jan 29, 2017
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#
#"and the factors are "(x-4),(x-8)# Hence the general equation is.
#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=-1toy=-(x^2-12x+32)=-x^2+12x-32#
graph{(y-x^2+12x-32)(y+x^2-12x+32)=0 [-10, 10, -5, 5]}
