# Thread: finding x and y intercepts of a parabola problem

1. ## finding x and y intercepts of a parabola problem

I cant see where im going wrong here can somebody point it out to me please. Il show my workings.
Trying to find the x-intercept
y= -2(x-3)^2 +8

so when y=o
0= -2(x+3)^2 +8
8= -2(x+3)^2
-4= (x+3)^2
-4= x^2 + 6x +9
x^2 + 6x + 13 now i dont know whats going on, if i use the formula -b +- (square root) b^2 -4ac /2a i get square root -16 which means x is not a real number? i think im going wrong somewhere on top.

any help appreciated my math is rusty

2. Originally Posted by dragon555
I cant see where im going wrong here can somebody point it out to me please. Il show my workings.
Trying to find the x-intercept
y= -2(x-3)^2 +8

so when y=o
0= -2(x+3)^2 +8
8= -2(x+3)^2
-4= (x+3)^2
-4= x^2 + 6x +9
x^2 + 6x + 13 now i dont know whats going on, if i use the formula -b +- (square root) b^2 -4ac /2a i get square root -16 which means x is not a real number? i think im going wrong somewhere on top.

any help appreciated my math is rusty
You don't need to expand to solve this problem and you also lost a minus sign on the 2nd line

$\displaystyle \displaystyle 0=-2(x+3)^2+8 \iff -8=-=2(x+3)^2 \iff 4=(x+3)^2$

Now we can just take the square root of both sides to get

$\displaystyle \displaystyle \pm 2 = (x+3) \iff -3 \pm 2=x$

So $\displaystyle x=-5;x=-1$

3. Originally Posted by dragon555
...which means x is not a real number? ...
As noted, you did make a mistake, but it's also possible that there be no x-intercepts (i.e. real solutions when y = 0).
In vertex form:
$\displaystyle y = a(x - h)^2 + k$

If "a" and "k" are both positive (negative), then the parabola will have no (real) x-intercepts.

4. ah i see what happed on the second line with the minuses. thanks for the help guys now i can draw the correct graph lol