# Quadratic "fun"ction

Printable View

• December 5th 2012, 05:41 PM
M670
Quadratic "fun"ction
So I got one problem perfect but this one is just not happening ....

Consider the Quadratic function http://webwork.mathstat.concordia.ca...75097b6b91.png.
Its vertex is (, ).
Its http://webwork.mathstat.concordia.ca...dd0b8b8e91.png-intercepts are http://webwork.mathstat.concordia.ca...9019052b81.png .
Note: If there is more than one answer enter them separated by commas.
Its http://webwork.mathstat.concordia.ca...22b824af51.png-intercept is http://webwork.mathstat.concordia.ca...b39caf0291.png .
So I am being told -46 and my x intercepts are wrong?
but to find $k= c-\frac{b^2}{4a}$ $k=-21-\frac{10^2}{4}$ and then when I plug the function into this $\frac{-b+-\sqrt b^2-4ac}{2a}$ I got $-5+-\sqrt46$ Sorry I cant seem to get my root to cover the whole part over b^2-4ac

My answer are vertex = (5,-46) $x=-5+\sqrt46$ $x=-5-\sqrt46$ and $y=-21$
• December 5th 2012, 05:46 PM
skeeter
Re: Quadratic "fun"ction
$f(x) = -x^2 + 10x - 21$

$f(x) = -(x^2 - 10x + 21)$

$f(x) = -(x - 3)(x - 7)$

try again ...
• December 5th 2012, 05:49 PM
M670
Re: Quadratic "fun"ction
Quote:

Originally Posted by skeeter
$f(x) = -x^2 + 10x - 21$

$f(x) = -(x^2 - 10x + 21)$

$f(x) = -(x - 3)(x - 7)$

try again ...

But what about if I do it this way
$x^2 + bx + c \;=\; (x-h)^2 + k,\quad\text{where}\quad h = -\frac{b}{2} \quad\text{and}\quad k = c - \frac{b^2}{4}.$
• December 5th 2012, 05:55 PM
HallsofIvy
Re: Quadratic "fun"ction
And after you have done that, what does it tell you about the x-intercepts? Also notice that your given function is NOT of the form " $x^2+ bx+ c$"
• December 5th 2012, 05:56 PM
skeeter
Re: Quadratic "fun"ction
$f(x) = -x^2 + 10x - 21$

$f(x) = -(x^2 - 10x) - 21$

$f(x) = -(x^2 - 10x + 25) - 21 + 25$

$f(x) = -(x-5)^2 + 4$
• December 5th 2012, 05:59 PM
M670
Re: Quadratic "fun"ction
Quote:

Originally Posted by HallsofIvy
And after you have done that, what does it tell you about the x-intercepts?

Oh I see what skeeter was trying to show me now, my x intercepts... But I still can't get the K value? and my other question why does the
$x=\frac{-b \pm \sqrt {b^2-4ac}}{2a},$ not give my x intercept?
• December 5th 2012, 06:00 PM
M670
Re: Quadratic "fun"ction
Does this question have to be solved in the manner you are showing me?
• December 5th 2012, 06:05 PM
M670
Re: Quadratic "fun"ction
So does this apply to all my quadratic equations with a coefficient of -1 infront of my x^2 I need to complete the square ? and not solve by using the quadratic formula?
• December 5th 2012, 07:23 PM
Recyclop
Re: Quadratic "fun"ction
for your vertex use (-b/2a, F( -b/2a)) those will be the coordinates for your vertex.
after that factor your quadratic as skeeter did above and set each binomial equal to zero and solve for x, those will be your roots(or x-intercepts).
for example: X^2-3x-4, when factored becomes (x-4)(x+1) set (x-4) =0 and solve do the same for (x+1). those would give you roots.
and it seems you already know how to do y.