What is the vertex and inverse of this parabola? What am I doing wrong? HELP!

meant to say
"...unless coefficient of x² is 1"
guess you figured that out.

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y = -x^2 + 4x + 0, x Vertex is always at x = -b/2a
x = 2 and at this point
y = -4 + 8 = +4

slightly different to your result
Vertex: (2, +4) opens down
Already given domain x Range: y is from +4 to minus infinity

Solve to make x subject of equation
x^2 - 4x + y = 0
x = 2 +/- sqrt(4 - y)
but other condition is x so x = 2 - sqrt(4 - y)
Reverse x, y roles to reveal inverse function
y = 2 - sqrt(4 - x)

This function stops at 4, 2
(because with say x = 5 you get imaginary quantity)

Domain: x: {minus infinity to plus infinity)
(All these inputs are allowed even though
outputs not defined for values of y > 2)
Range: is from 2 to minus infinity

Graphing them both on the same plane:
You already know that the first function
is an inverted parabola peaking at (2, 4)
The second function gently curves up
from (-5, -1) through the origin and stops
at (4, 2) on the parabola
(so intersections look a bit like a fingernail)
Plot a few more points if you want to.
Hope that helps,

Regards - Ian

Your computation of the vertex appears to have an error. The vertex of
.. ax^2 + bx + c
is given by
.. x = -b/(2a)

You have b=4, a=-1, so the x value (as you have calculated) is
.. x = -4/(2(-1)) = 2
The y value is
.. y = 4(2) - (2)^2 = 8 - 4 = 4

The vertex is (2, 4). The domain of x is (-Infinity, 2] by definition, and the range of y will be (-Infinity, 4].

Solving the equation for x, we get
.. 4 - y = x^2 - 4x + 4 ... negate and add 4
.. √(4-y) = ±(x-2) ... square root both sides
.. x = 2 - √(4-y) ... choose the branch of the function of interest.

The domain of this is the same as the range of the original: (-Infinity, 4]. The range of this is the same as the domain of the original: (-Infinity, 2]. The inverse function is not defined for values of y > 4.