Find the domain and range for each quadratic function.

a. y = x^2 – 6x +5

b. y = 3x^2 + 2x +9

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- Aug 2nd 2006, 07:53 PMpashahDomain and Range
Find the domain and range for each quadratic function.

a. y = x^2 – 6x +5

b. y = 3x^2 + 2x +9 - Aug 2nd 2006, 10:29 PMCaptainBlackQuote:

Originally 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 - Aug 3rd 2006, 07:30 AMThePerfectHackerQuote:

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,

$\displaystyle y\geq -4$

Quote:

b. y = 3x^2 + 2x +9

$\displaystyle 3x^2+2x+(9-y)=0$

Work with discrimanant,

$\displaystyle (2)^2-4(3)(9-y)\geq 0$

$\displaystyle -104+12y\geq0$

$\displaystyle 12y\geq 104$

$\displaystyle y\geq 8.\bar 6$