• February 19th 2013, 05:17 PM
Civy71
Find the quadratic polynomial whose graph passes through the points (0,0) (-1,1) and (1,1).
• February 19th 2013, 05:34 PM
jakncoke
$x^2$
• February 19th 2013, 05:42 PM
Civy71
Thanks Dave, how did you arrive at that answer?
• February 19th 2013, 05:46 PM
jakncoke
I used solved the linear system 0 = f(0) = $a_3$ 1 = f(-1) = $a_2 - a_1 + a_0$ 1 = f(1) = $a_2 + a_1 + a_0$.
Then i looked at the question, the points and slapped my self on my forehead and erased all my work to save embarrassment.
• February 19th 2013, 06:59 PM
HallsofIvy
Any "quadratic polynomial" can be written in the form $y= ax^2+ bx+ c$ for some numbers, a, b, and c. When x= 0, y= 0 so $0= a(0^2)+ b(0)+ c$ so c= 0. Then when x= -1, y= 1 so $1= a(-1)^2+ b(-1)$ or $a- b= 1$. Finally, when x= 1, y= 1 so $1= a(1)^2+ b(1)$ or a+ b= 1.