i don't know if i did this right...
sketch 2x^2 +12x -14 and indicate all its vertex and intercepts
my vertex tht i got is (-3,-26)
my y-intercept is (0,-14) x-intercept (-1,0) and (7,0)
is this right?
Hi there.
One method of finding the minimum of the graph is to convert it to turning point form by completing the square:
$\displaystyle y=2x^2-12x-14$
$\displaystyle y=2(x^2+6x-7)$
$\displaystyle y=2((x+3)^2-16)$
$\displaystyle y=2(x+3)^2-32$
You can now see that the minimum is at $\displaystyle (-3,-32)$
Now to find the x-intercepts let $\displaystyle y=0$
$\displaystyle 2(x+3)^2-32=0$
$\displaystyle (x+3)^2-16=0$
$\displaystyle (x+7)(x-1)=0$
Therefore the x-axis intercepts are at $\displaystyle (-7,0)$ and $\displaystyle (1,0)$
The y-intercept occurs where $\displaystyle x=0$ which is at $\displaystyle (0,-14)$
You should post your working so we can see where you went wrong.