Question

How do I find the domain and range in this quadratic: y=x^2-2x+4?

Answer 1

For polynomials, the domain is always going to be \([-\infty, \infty]\)

Answer 2

For quadratics, to find the range you find the vertex and figure out if it opens up or down.

Answer 3

So how do I write the domain exactly ?

Answer 4

\[
\begin{split}
y &=x^2-2x+4 \\
&= (x-1)^2+3 \\
y-3 &= (x-1)^2

\end{split}
\]This means the vertex is at \((1, 3)\)

Answer 5

To write the domain you'd just write \((-\infty , \infty ) \)

Answer 6

Or you could say "all real numbers"

Answer 7

Oh How do you know it's all real nimbers?

Answer 8

For these types of problems, we generally ignore complex numbers.
The reason we know it is all real numbers is because we never divide by 0 or take the square root of a negative number.

Answer 9

Okay, Thanks (:

Answer 10

Is it always all real numbers for the domain ?

Answer 11

For polynomials, yes.
For a function like \(f(x)=1/x\), no.

Answer 12

Would the range be y is -1 ?

Answer 13

Or greater than or equal to or less than or equal to ?

Answer 14

It's opening up, so it can't go any lower than the vertex.

Answer 15

where did the \(-1\) come from?

Answer 16

On one of my older papers they switched the signs.

Answer 17

first coordinate of the vertex is always \(-\frac{b}{a}\) in your example it is
\[-\frac{-2}{2\times 1}=1\] the second coordinate is what you get when you replace \(x\) by \(1\), in this case you get \(y=1^2-2\times 1+4=3\)

Answer 18

But what confusing me is I don't know how to write it. The range.

Answer 19

since \(3\) is the lowest the curve can be, it is \([3,\infty)\)