!write a system of linear inequalities to each graph. need to do both of them

The system of linear inequalities to each graph is the following

1) (Largest graph)

(1) y≤-3x

(2) y≥2x-4

2) (Smallest graph)

(1) y≤x+2

(2) y≤-x-3

Step-by-step explanation:

1) Largest graph

We can see two right lines

a) One of the lines (Line 1) goes through:

Origin: O=P1=(0,0)=(x1, y1)→x1=0, y1=0

P2=(2, -6)→x2=2, y2=-6

Slope line 1: m1=(y2-y1)/(x2-x1)

Replacing the known values:

m1=(-6-0)/(2-0)

m1=(-6)/(2)

m1=-3

Equation line 1:

y-y1=m1 (x-x1)

y-0=-3(x-0)

y=-3x

The region is below this right line, then the first inequality is:

(1) y≤-3x

b) The other line (Line 2) goes through:

P3=(2,0)=(x3, y3)→x3=2, y3=0

P4=(5, 6)→x4=5, y4=6

Slope line 2: m2=(y4-y3)/(x4-x3)

Replacing the known values:

m2=(6-0)/(5-2)

m2=(6)/(3)

m2=2

Equation line 2:

y-y3=m2 (x-x1)

y-0=2(x-2)

y=2x-4

The region is above this right line, then the second inequality is:

(2) y≥2x-4

Then, the system of linear inequalities for the largest graph is:

(1) y≤-3x

(2) y≥2x-4

2) Smallest graph

We can see two right lines

a) One of the lines (Line 1) goes through:

P1=(0, 2)=(x1, y1)→x1=0, y1=2

P2=(5, 7)→x2=5, y2=7

Slope line 1: m1=(y2-y1)/(x2-x1)

Replacing the known values:

m1=(7-2)/(5-0)

m1=(5)/(5)

m1=1

Equation line 1:

y-y1=m1 (x-x1)

y-2=1(x-0)

y-2=1(x)

y-2=x

y-2+2=x+2

y=x+2

The region is below this right line, then the first inequality is:

(1) y≤x+2

b) The other line (Line 2) goes through:

P3=(0,-3)=(x3, y3)→x3=0, y3=-3

P4=(5, -8)→x4=5, y4=-8

Slope line 2: m2=(y4-y3)/(x4-x3)

Replacing the known values:

m2=(-8-(-3))/(5-0)

m2=(-8+3)/(5)

m2=(-5)/5

m2=-1

Equation line 2:

y-y3=m2 (x-x1)

y-(-3)=-1(x-0)

y+3=-1(x)

y+3=-x

y+3-3=-x-3

y=-x-3

The region is below this right line, then the second inequality is:

(2) y≤-x-3

Then, the system of linear inequalities for the smallest graph is:

(1) y≤x+2

(2) y≤-x-3

207/8 is the right

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.