Question

Question 2.4
Kindly help me with this question..This isn't homework, i am just trying to work through problems as i am helping my son with his maths as he is disadvantage.

I have done most of the questions but i am stuck with question 2.4. So the assistance would really be appreciated...

I have attached a picture regarding the question as the link doesn't work.

You know from solving question 2.1 that the vertex of the parabola is D(-1, 4).

Find the distance d between A(-3, 0) and D(-1, 4):
rise = 4 - 0 = 4
run = -1 - (-3) = 2

By the Pythagorean Theorem,
d² = rise² + run² = 4² + 2² = 20
d = √20 = 2√5
:::::
Equation for circle of radius r, centered at (h, k):
(x-h)² + (y-k)² = r²

So the equation for the circle of radius 2√5, centered at A(-3, 0) is:
(x+3)² + y² = 20
:::::
Question 2.6
You already know from solving questions 2.1 and 2.2 that
f(x) = -x² - 2x + 3
g(x) = 2x + 6

Let u(x) = f(x) - g(x) = -x² - 4x - 3.
When u(x) is a maximum, u'(x) = 0
u'(x) = -2x - 4 = 0
x = -2
u(-2) = -(-2)² - 4(-2) - 3 = 1
The maximum vertical distance between f and g is 1.

Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Thank you soooo much for taking time out to assist me with this question. I really truly appreciated it.

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#### Janet Yang

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