Question

1. find the values of a, b, and c for $y = x^2 - 9x + 18$.

$a = $
$b = $
$c = $

1. find the real roots for $y = x^2 - 9x + 18$. use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

options: $(-9, 18)$, $(0, -6)$, $(6, 0)$, $(0, -3)$, $(3, 0)$

Explanation:

Step1: Identify quadratic coefficients

For $y = ax^2 + bx + c$, compare with $y = x^2 - 9x + 18$:
$a=1$, $b=-9$, $c=18$

Step2: Substitute into quadratic formula

$$x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(18)}}{2(1)}$$

Step3: Calculate discriminant

$$\sqrt{81 - 72} = \sqrt{9} = 3$$

Step4: Solve for two roots

First root: $x = \frac{9 + 3}{2} = 6$
Second root: $x = \frac{9 - 3}{2} = 3$
These correspond to points $(6,0)$ and $(3,0)$.

Answer:

1. $a=1$, $b=-9$, $c=18$
2. $(6,0)$, $(3,0)$