Y-y1=m(x-x1)
for slope=m
a point =(x1,y1)
slope=2
point= (-1,3)
y-3=2(x+1)
y-3=2x+2
y=2x+5
test given points
A.
(1,5)
5=2(1)+5
5=7
false
B.
(1,4)
4=2(1)+5
4=7
false
C.
(-3,2)
2=2(-3)+5
2=-6+5
2=-1
false
D.
(0,5)
5=2(0)+5
5=0+5
5=5
true
answer is D
Answer: The range is

The way to get this answer isn't completely obvious. What I did first was graph the function using GeoGebra allowing k to be a parameter. With k as a sliding parameter, the graph was able to be slid up and down the screen. The basic shape of the curve is a distorted W shape. The lowest point being the local minimum. Which is where the lower endpoint of the range would be.
Using the derivative, we can get
y = x^4 + 8x^3 + k
dy/dx = 4x^3 + 24x^2
set dy/dx equal to zero and solve for x
dy/dx = 0
4x^3 + 24x^2 = 0
4x^2(x + 6) = 0
4x^2 = 0 or x+6 = 0
x = 0 or x = -6
It turns out that x = 0 is a saddle point so we can ignore this value. Only x = -6 is where the local min is at. Plug this back into the original function
y = x^4+8x^3+k
y = (-6)^4+8(-6)^3+k
y = -432+k
y = k - 432
So when the input is x = -6, the output is k - 432 for some value k
In other words, the smallest y can get is k - 432. Otherwise it will be larger than this value
So that's why the range is

Note: if this calculus approach doesn't make much sense, then you can try various k values and use a calculator to find the lowest point. Making a table of k values with the corresponding lowest point would hopefully point you in the right direction.
Answer:
<em>Translater</em>
Step-by-step explanation:
<em>F = (4x-3) squared - (x + 3) (3-9x)</em>
<em>- develop and reduce (4x-3) to the edge</em>
<em>- show that F = (5x) squared</em>
<em />
- <em>(NOT AN ANSWER ITS TO HELP OTHERS WHO DOESN"T KNOW FRENCH)</em>
Is there a picture that suppose to come with this
Answer:
m=-2
Step-by-step explanation:
Take 2 points (0,0) and (-3,6)
m(slope)= m = rise/run = y2 - y1/ x2 - x1
y2-y1= 6-0=6
x2-x1 = -3-0 = -3
m=6/-3 = -2
y=-2x