2.6 Question 1

[Eigenvalue]

Precalculus
Michael Sullivan
Edition 4
Chapter 2, Section 2.6

Let P=(x,y) be a point on the graph of y=x²-8.

a) Express the distance of P to the origin as a function of x

b) What is d if x=0

c) What is d if x=1

d) Use a graphing utility to graph y=d(x)

e) For what values of x is d smallest

[nycmath]

Part A


Let (0, 0) = origin

We are given the point P to be (x, y).

Notice that y = [tex]x^{2 }-8[/tex].

Replace y with [tex]x^{2 }-8[/tex].

Let d = distance

d = [tex]\sqrt{(x)^2 + (x^2-8)^2}[/tex]

d(x) = [tex]\sqrt{x^4-15x^2+64}[/tex]

[nycmath]

Part B

Let x = 0

d(1) = [tex]\sqrt{(0)^2-15(0)^2+64}[/tex]

d(1) = [tex]\sqrt{64}[/tex]

d(1) = 8

[nycmath]

Part C

Let x = 1

d(1) = [tex]\sqrt{(1)^2-15(1)^2+64}[/tex]

d(1) = [tex]\sqrt{1-15+64}[/tex]

d(1) = [tex]\sqrt{-14+64}[/tex]

d(1) = [tex]\sqrt{50}[/tex]

d(1) = [tex]\sqrt{25}[/tex]•[tex]\sqrt{2}[/tex]

d(1) = 5[tex]\sqrt{2}[/tex]

[Eigenvalue]

Part E

Find the minimum value for x⁴-15x²+64
Set x²=u
u²-15u+64
The vertex is [tex]\frac{15}{2}[/tex]
Substitute [tex]\frac{15}{2}[/tex] into the equation

([tex]\frac{15}{2}[/tex])²-15([tex]\frac{15}{2}[/tex])+64=[tex]\frac{31}{4}[/tex]

As the expression is under a square root, the minimum values are [tex]\pm[/tex] [tex]\frac{ \sqrt{31} }{2}[/tex]

[nycmath]

Part E

Find the minimum value for x⁴-15x²+64
Set x²=u
u²-15u+64
The vertex is [tex]\frac{15}{2}[/tex]
Substitute [tex]\frac{15}{2}[/tex] into the equation

([tex]\frac{15}{2}[/tex])²-15([tex]\frac{15}{2}[/tex])+64=[tex]\frac{31}{4}[/tex]

As the expression is under a square root, the minimum values are [tex]\pm[/tex] [tex]\frac{ \sqrt{31} }{2}[/tex]

Great work but you didn't let me finish the problem.

[nycmath]

Part D

Let y = d(x)

Here is the graph of y = [tex]\sqrt{x^4-15x^2+64}[/tex]