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Question

Let P(x, y) be a point on the graph of y = x^2 - 2.

a. Write the domain and range for the function in interval notation: D = _____, R= ________

b. What is the coordinate of the minimum point on the function? Minimum = _________ at ___________.

c. Is this function even or odd? Explain. Yep

d. Write the inverse function f^-1 (x) = ______.

e. Express the distance, d, from P to the origin (0,0) as a function of the point's x-coordinate. d= ________________ (Hint: Distance formula = √(𝑥2 − 𝑥1 )^2 + (𝑦2 − 𝑦1 )^2

Answer 1

y= x^2 -2 is an upward opening parabola where the vertex is the minimum point

vertex is (0, -2)

domain is any real number, range is y>-3

it's an even function f(-x) = -f(x). It's symmetric about the y axis

to find the inverse function switch x and y

x=y^2-2

and solve for the new y

y = +/- sqr(x+2)

but that's not a function, it's the inverse relation. there is no inverse function

there is an inverse, but it's just not a function. It fails the vertical line test for a function

unless you use only the top or bottom half of the rightward opening parabola

drop either the + or - square root

for y=x^2 -2

the distance from a point on the parabola, the original function, (x,y) to the origin (0,0) is

d= sqr(x^2 +y^2)

d= sqr(x^2 + (x^2 -2)^2)

d = sqr(x^4 -3x^2 +4)

distance is generally positive so ignore the negative square root

and for the distance from (x,y) to the origin for the inverse relation

d = sqr(x^2 + (sqr(x+2)^2)

d= sqr(x^2 +x +2)