# Math Help - Quadratic "fun"ction

So I got one problem perfect but this one is just not happening ....

Its vertex is (, ).
Its -intercepts are .
Note: If there is more than one answer enter them separated by commas.
Its -intercept is .
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 $y=-21$

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

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

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

try again ...

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
$x^2 + bx + c \;=\; (x-h)^2 + k,\quad\text{where}\quad h = -\frac{b}{2} \quad\text{and}\quad k = c - \frac{b^2}{4}.$

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 " $x^2+ bx+ c$"

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

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

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

$f(x) = -(x-5)^2 + 4$

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?

Does this question have to be solved in the manner you are showing me?