whose zeros are
a) 3 and 5
b) –4 and 3
c) ½ and ⅔
d) 2 ± √3
A quadratic function with zeros at a and b is:
y = (x - a)(x - b)
Then we can multiply out the right side of that equation, or "FOIL" it, to get:
y = x2 - bx - ax + ab
And then continue to simplify the right side until it looks like a nice quadratic equation.
Using this "template", we can find a solution to each of these problems.
a) y = (x - 3)(x - 5) = x2 - 5x - 3x + 15 = x2 - 8x + 15
b) y = (x + 4)(x - 3) = x2 - 3x + 4x - 12 = x2 + x - 12
c) y = \((x-\frac12)(x-\frac23)\ =\ x^2-\frac23x-\frac12x+\frac13\ =\ x^2-\frac76x+\frac13\)
d) In this case, the one root is 2 + √3 and the other is 2 - √3 , so the quadratic equation is:
y = ( x - (2 + √3) )( x - (2 - √3) )
y = ( x - 2 - √3 )( x - 2 + √3 )
y = x2 - 4x + 1
Can you figure out the last one? It might be the trickiest one, so if you need more help on it just ask!
Here is a graph to check the answers: https://www.desmos.com/calculator/lv66bdc1jy