1. parabola

Place the following in standard quadratic form. (which is (x-h)+k ). The find the vertex, axis, domain, range, and zeroes.
(can be all done with a calculator, no work needed.)

1. f(x) = x^2-8x+16
2. f(x)=16-x^2
3. f(x)=x^2+4x+5
4. f(x)= -x^2+2x+5
5. f(x)= 4x^2-4x+21
6. f(x)= 2x^2-x+1

ANYONE?

2. Originally Posted by iloveyouxD
Place the following in standard quadratic form. (which is (x-h)+k ). The find the vertex, axis, domain, range, and zeroes.
(can be all done with a calculator, no work needed.)

1. f(x) = x^2-8x+16
2. f(x)=16-x^2
3. f(x)=x^2+4x+5
4. f(x)= -x^2+2x+5
5. f(x)= 4x^2-4x+21
6. f(x)= 2x^2-x+1
I only have time to do the 1st one. Post each function individually and you may get a faster response. Good luck.

(1) $f(x)=x^2-8x+16$

$f(x)=(x-4)^2+0$

$V(4, 0)$

Axis of symmetry $x=4$

Domain $D=\{x|x \in \Re \}$

Range $R=\{y|y \ge 0\}$

Zeros $x= 4$ "double root"

3. Originally Posted by iloveyouxD
Place the following in standard quadratic form. (which is (x-h)+k ). The find the vertex, axis, domain, range, and zeroes.
(can be all done with a calculator, no work needed.)

1. f(x) = x^2-8x+16
2. f(x)=16-x^2
3. f(x)=x^2+4x+5
4. f(x)= -x^2+2x+5
5. f(x)= 4x^2-4x+21
6. f(x)= 2x^2-x+1

ANYONE?
3.
$f(x) = x^2+4x+5$
$f(x) = (x^2+4x)+5$
$f(x) = (x^2+4x+4-4)+5$
$f(x) = (x+2)^2 +1$
Vertex $(-2,1)$
Domain, same as masters solution to 1. x is an element of real number
Range same as masters except it is 1
Axis=-2

Follow the same approach for the others.

4. The standard form is f(x)= ax^2+bx+c . To find the vertex find x-coordinate first using the formula x=-b/2a. The axis is the vertical line with the same x coordinate. Domain is all possible numbers of x you may use, Range is a set of all possible y you are getting by each given formula.To find 0's factor each equation and make it =0, solve it.
I'll do #2 f(x)=16-x^2. Lets rewrite it in standard form -x^2+16, so a=-1, b=0, c=16.
Vertex: x= -b/2a= 0. Plug x=o into y=16-x^2 to find y coordinate of the vertex:y=16, so vertex coordinates are(0,16)
Axis: vertical line x=0(x-axis itself)
Domain: all x's , or x is any real number.
Range is also all real numbers( in #1you would never get negative answer, that's why y>=0)
Zeros: Let's factor it first, -x^2+16=-1(x^2-16)= -1(x-4)(x+4). Make it =0 and solve it, x1=4, x2=-4, so zeros are (4,0), (-4,)