Solving a Parabola Question: Help Appreciated

[DethRose]

Im In a 1st Year University math class and its been a while since i did this stuff and can't remember if I am 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 don't 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

 

[Dick]

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

 

[DethRose]

great thanks for the help!

 

[DethRose]

was trying to graph this and isn't 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?

 

[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.

 

[DethRose]

that makes sense i think there's 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

 

[HallsofIvy]

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!)