Parabola Question

1. Sep 18, 2007

DethRose

Im In a 1st Year University math class and its been a while since i did this stuff and cant remember if im doing this question right so any help would be much appreciated.

The question is to find all the info from the equation y=2(x-4)^2-3

I used the b+/- b^2 -4ac... equation and ended up getting 16+/- square root of 24/4.

This gives an odd number (5.22474, 2.77525) so i dont think it is right. When i apart the original equation to make into ax^2+bx+c=0 form i did 2(x^2-8x+16)-3=0 and i am not sure if that is correct either or if it should be multiplies without brackets.

Thanks for any help

2. Sep 18, 2007

Dick

If your problem is to find the x-intercepts (roots of the quadratic), then you did it correctly

3. Sep 18, 2007

DethRose

great thanks for the help!

4. Sep 18, 2007

DethRose

was trying to graph this and isnt the vertex (-4,-3), so then how could those x intercepts be right since they are both on positive side of the x axis, and the vertex is in the negative side?

5. Sep 18, 2007

Dick

Because you didn't get the vertex right. It should be at the minimum value of y, right? x=-4 is not the minimum, try x=+4.

6. Sep 18, 2007

DethRose

that makes sense i think theres a typo on the assignment then cause it says (p,q) are the coordinates of the vertex in the form of y=a(x-p)^2+q

and then it gives y=2(x-4)^2-3

7. Sep 18, 2007

HallsofIvy

Staff Emeritus
Why would you think there is a typo? (p,q) is the vertex of y= a(x-p)2+ q because when x= p, y= a(p-p)2+ q= a(0)+ q= q while for any other value of x, x-p is non-zero, (x-p)2 is positive so y= a(x-p)2+q is q plus some positive number: (p, q) is the lowest point on the graph.

Comparing the general for m y= a(x-p)2+ q with y= 2(x-4)2-3 isn't it clear that p= 4 and q= -3?
(Notice that it is (x-p)2 and (x-4)2, not (x-(-4))2!)