• Dec 5th 2012, 05:41 PM
M670
So I got one problem perfect but this one is just not happening ....

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 $y=-21$
• Dec 5th 2012, 05:46 PM
skeeter
$f(x) = -x^2 + 10x - 21$

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

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

try again ...
• Dec 5th 2012, 05:49 PM
M670
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
$x^2 + bx + c \;=\; (x-h)^2 + k,\quad\text{where}\quad h = -\frac{b}{2} \quad\text{and}\quad k = c - \frac{b^2}{4}.$
• Dec 5th 2012, 05:55 PM
HallsofIvy
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 " $x^2+ bx+ c$"
• Dec 5th 2012, 05:56 PM
skeeter
$f(x) = -x^2 + 10x - 21$

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

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

$f(x) = -(x-5)^2 + 4$
• Dec 5th 2012, 05:59 PM
M670
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?
• Dec 5th 2012, 06:00 PM
M670