Math

QuestionFind the starting point of the function f(x)=x4+7f(x)=\sqrt{x-4}+7 on the coordinate plane. What is the ordered pair?

Studdy Solution

STEP 1

Assumptions1. The function is f(x)=x4+7f(x)=\sqrt{x-4}+7 . We are looking for the starting point of the function on the coordinate plane

STEP 2

The square root function, x\sqrt{x}, is defined for all x0x \geq0. In the function f(x)=x4+7f(x)=\sqrt{x-4}+7, the expression inside the square root is x4x-4. This means that x4x-4 must be greater than or equal to0 for the function to be defined.
x40x-4 \geq0

STEP 3

To find the value of xx that makes x0x- \geq0, we can solve this inequality.
xx \geq

STEP 4

This means that the function f(x)=x4+7f(x)=\sqrt{x-4}+7 starts at x=4x=4 on the coordinate plane.

STEP 5

Now, we need to find the corresponding yy-value when x=4x=4. We can do this by substituting x=4x=4 into the function.
f(4)=44+7f(4)=\sqrt{4-4}+7

STEP 6

implify the expression inside the square root.
f(4)=0+f(4)=\sqrt{0}+

STEP 7

Calculate the square root of0.
f(4)=0+7f(4)=0+7

STEP 8

Calculate the value of f(4)f(4).
f(4)=7f(4)=7So, the function f(x)=x4+7f(x)=\sqrt{x-4}+7 begins at the ordered pair (4,7)(4,7) on the coordinate plane.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord