Find the domain and range for each quadratic function.
a. y = x^2 – 6x +5
b. y = 3x^2 + 2x +9
The RHS of both of these are defined for all of $\displaystyle \mathbb{R}$, soOriginally Posted by pashah
their domains are both the set of all real numbers.
Because these both represent parabolas which go to $\displaystyle \infty$ as $\displaystyle x$
goes to $\displaystyle \pm \infty$, they can take all values greater than teir minimums.
The minimum of the first occurs when $\displaystyle 2x-6=0$, which is at
$\displaystyle x=3$, where $\displaystyle y=-4$, so for a. the range is $\displaystyle y \ge 4$.
The minimum of the second occurs when $\displaystyle 6x+2=0$, which is at
$\displaystyle x=-1/3$, where $\displaystyle y=26/3$, so for a. the range is $\displaystyle y \ge 26/3$.
RonL
I do this slight diffrently.Originally Posted by pashah
Rewrite as,
$\displaystyle x^2-6x+(5-y)=0$
Now, you need to find all $\displaystyle y$ such as this quadradic equation has a real solution. That happens when, the discrimanant is non-negative.
$\displaystyle (-6)^2-4(1)(5-y)\geq 0$
Simplify,
$\displaystyle 16+4y\geq 0$
Thus,
$\displaystyle 4y\geq -16$
Thus,
$\displaystyle y\geq -4$
Proceede the same,b. y = 3x^2 + 2x +9
$\displaystyle 3x^2+2x+(9-y)=0$
Work with discrimanant,
$\displaystyle (2)^2-4(3)(9-y)\geq 0$
$\displaystyle -104+12y\geq0$
$\displaystyle 12y\geq 104$
$\displaystyle y\geq 8.\bar 6$