Question

Given the function f(x) = -5(x+4)^2 +15 complete the following. Please show all of your work.
i. If f(x) = 15, what is x?
ii. What point(s) from part (i) are on the graph of f?
iii. List the x-intercepts, if any, of the graph of f.
iv. List the y-intercept, if there is one, of the graph of f.
Given the function f(x) = -5(x+4)^2 +15 complete the following. Please show all of your work.
i. If f(x) = 15, what is x?
ii. What point(s) from part (i) are on the graph of f?
iii. List the x-intercepts, if any, of the graph of f.
iv. List the y-intercept, if there is one, of the graph of f.
@Mathematics

Answer 1

i.


f(x) = -5(x+4)^2 +15


15 = -5(x+4)^2 +15


15-15 = -5(x+4)^2 +15-15


0 = -5(x+4)^2


-5(x+4)^2 = 0


(x+4)^2 = 0/(-5)


(x+4)^2 = 0


x+4 = sqrt(0)


x+4 = 0


x = -4

Answer 2

ii.

Since f(x) = 15 and y is the same as f(x), this means that y = 15


But we know that when f(x) = 15, then x = -4


So when x = -4, y = 15


So the point (-4, 15) lies on the graph

Answer 3

iii.

f(x) = -5(x+4)^2 +15


0 = -5(x+4)^2 +15


-15 = -5(x+4)^2


-15/(-5) = (x+4)^2


3 = (x+4)^2


(x+4)^2 = 3


x+4 = +-sqrt(3)


x+4 = sqrt(3) or x+4 = -sqrt(3)


x= -4 + sqrt(3) or x = -4 -sqrt(3)


x = -2.2679 or x = -5.7321


So the x-intercepts are approximately (-2.2679 , 0) and (-5.7320, 0)

Answer 4

iv.


f(x) = -5(x+4)^2 +15


f(0) = -5(0+4)^2 +15


f(0) = -5(4)^2 +15


f(0) = -5(16) +15


f(0) = -80 +15


f(0) = -65


So the y-intercept is (0, -65)

Answer 5

Wow! Awesome. Thanks!!

Answer 6

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