Question

Which of the following are quadratic functions? Check all that apply.

A. y=5+2x
B. y=2(x-3x^2+1)
C. y=1/2x^2-13x+4
D. y=x/x-1

Answer 1

@KeithAfasCalcLover do you mind helping with this?

Answer 2

This is the quadratic formula -
x = [ -b ± sqrt(b^2 - 4ac) ] / 2a

Answer 3

Sure Ali :)
Im just going to make 'em a little neat:
We have:
\[\eqalign{
&A.\phantom{a}y=5+2x \\
&B.\phantom{a}y=2(x-3x^2+1) \\
&C.\phantom{a}y=\frac{x^2}{2}-13x+4 \\
&D.\phantom{a}y=\frac{x}{x-1} \\
}\]
Right?

Answer 4

yes sir! :)

Answer 5

Beautiful.
So then the criteria for a quadratic I would say...is:
-It has to be a second degree polynomial (meaning it has to have the "\(x^2\)" part lol
-And it has to be re-arrangeable to the form \(ax^2+bx+c\). Where \(\{a,b,c\in R|a,b,c\neq0\}\). That means that the co-efficients have to be regular numbers like \(5.344\) or \(-6\frac{3}{4}\) and cant be zero.

So lets look at the first one:
Well the co-efficients are real numbers buuuut...it's missing the "\(x^2\)" part right?

Answer 6

Correct, so A would be out of the picture as well as D?

Answer 7

Nice job Ali. Exactly. So apparently, nothing is wrong with C and D. They both have an "\(x^2\)" and their \(a,b,\) and \(c\) are all real numbers. So I guess that's it lol

Answer 8

So it's C and D, not B and C?

Answer 9

Ehh, well its B and C remember? Cause D is not valid

Answer 10

That's what I thought but earlier you had put "nothing is wrong with C and D" so I got confused :o