321. I need to find the vertex of this parabola, as well as the domain and range. I also need to find its inverse....
y = 4x  x^2, given that x
This is what I have so far....
4 = x^2 + 4x +4
4 = (x+2)^2 (completing the square
0 = 1(x2)^2  4 (putting into standard form)
Vertex: (2, 4)
Opens down
y intercept: 8 (solving for x = 0)
Domain: x
Now, for the inverse.... this is what I have....
x = 1(y2)^2  4
x = (y2)^2 + 4 (multiplying by 1 on both sides)
x  4 = (y2)^2 (moving 4)
sqrt(x  4) = y  2 (rooting both sides)
OK, but here I am stuck because I have a square root of a negative number and I know that this problem does not deal with that stuff.... well, at least I don't think (or hope?) so.... Can you help me?
I need to find the inverse function, and the domain/range of the inverse? Also, I need to graph them both on the same plane, but you cant show me that on Y!A,.... so if you can help, please help me. Thanks!
y = 4x  x^2, given that x
This is what I have so far....
4 = x^2 + 4x +4
4 = (x+2)^2 (completing the square
0 = 1(x2)^2  4 (putting into standard form)
Vertex: (2, 4)
Opens down
y intercept: 8 (solving for x = 0)
Domain: x
Now, for the inverse.... this is what I have....
x = 1(y2)^2  4
x = (y2)^2 + 4 (multiplying by 1 on both sides)
x  4 = (y2)^2 (moving 4)
sqrt(x  4) = y  2 (rooting both sides)
OK, but here I am stuck because I have a square root of a negative number and I know that this problem does not deal with that stuff.... well, at least I don't think (or hope?) so.... Can you help me?
I need to find the inverse function, and the domain/range of the inverse? Also, I need to graph them both on the same plane, but you cant show me that on Y!A,.... so if you can help, please help me. Thanks!

it's hard to properly complete the square
unless coefficient of x = 1
so
y = x² + 4x
y = x²  4x
y + 4 = x²  4x + (2)²
y + 4 = (x  2)²
y = (x  2)²  4
y = (x  2)² + 4 ← see, you got the signs wrong
so the vertex is actually (2, 4)
does open down
or, in other words, vertex is max
so
range is y ≤ 4
and you were given the domain to start with
for the inverse,
I prefer to swap vars last, so
y = (x  2)² + 4
4  y = (x  2)²
±√(4  y) = x  2
2 ± √(4  y) = x
now, since you started with only the left side of the parabola,
you should end up with only the bottom half in its inverse:
2  √(4  y) = x
so
y = 2  √(4  x)
for the inverse domain and range,
just swap the domain and range of the original function:
D: x ≤ 4
R: y ≤ 2
see
http://s584.photobucket.com/albums/ss282…
unless coefficient of x = 1
so
y = x² + 4x
y = x²  4x
y + 4 = x²  4x + (2)²
y + 4 = (x  2)²
y = (x  2)²  4
y = (x  2)² + 4 ← see, you got the signs wrong
so the vertex is actually (2, 4)
does open down
or, in other words, vertex is max
so
range is y ≤ 4
and you were given the domain to start with
for the inverse,
I prefer to swap vars last, so
y = (x  2)² + 4
4  y = (x  2)²
±√(4  y) = x  2
2 ± √(4  y) = x
now, since you started with only the left side of the parabola,
you should end up with only the bottom half in its inverse:
2  √(4  y) = x
so
y = 2  √(4  x)
for the inverse domain and range,
just swap the domain and range of the original function:
D: x ≤ 4
R: y ≤ 2
see
http://s584.photobucket.com/albums/ss282…
12
keywords: and,inverse,of,this,vertex,is,parabola,am,wrong,doing,HELP,What,the,What is the vertex and inverse of this parabola? What am I doing wrong? HELP!