Question 1

Find all fixed points of the function. $f ( x ) = 13 + \sqrt { x - 1 }$
A) x = 17
B) x = 0, x = 170
C) x = 10
D) x = 17, x = 10
E) no fixed points

Accepted Answer

A fixed point is a value of x such that f(x) = x. 
Substituting f(x) into the equation, we get: 
f(x) = 13 + √(x-1) = x

Subtracting 13 from both sides: 
√(x-1) = x-13

Squaring both sides: 
x-1 = x^2 - 26x + 169

Moving all the terms to one side: 
x^2 - 27x + 170 = 0

Factoring the quadratic: 
(x-17)(x-10) = 0

Thus, the fixed points are x = 17 and x = 10. However, substituting x = 10 into the original equation, we get f(10) = 13 + √(9) = 16, which is not equal to 10. Therefore, x = 10 is not a fixed point. The only fixed point is x = 17.

Therefore, the best choice is A) x = 17.

Question 2

Suppose that the revenue generated by selling x units of a certain commodity is given by $R = - \frac { 1 } { 5 } x ^ { 2 } + 600 x$ . Assume that R is in dollars. What is the maximum revenue possible in this situation?
A) $450,000
B) $440,000
C) $430,000
D) $900,000
E) $480,000

Accepted Answer

The revenue function $R = -\frac{1}{5}x^2 + 600x$ is a quadratic equation in the form of $ax^2 + bx + c$, where $a = -\frac{1}{5}$, $b = 600$, and $c = 0$. The maximum revenue occurs at the vertex of the parabola represented by this equation. The x-coordinate of the vertex can be found using the formula $-\frac{b}{2a}$. Substituting the values of $a$ and $b$ gives $-\frac{600}{2(-\frac{1}{5})} = 1500$. Substituting $x = 1500$ back into the revenue equation to find the maximum revenue: $R = -\frac{1}{5}(1500)^2 + 600(1500) = -450,000 + 900,000 = 450,000$.

Question 3

Sketch the graph of the function and specify all x- and y-intercepts. y = - 3x⁴ + 5 A) x-intercepts: 5 Y-intercept: - 50,625 B) x-intercepts: -5 Y-intercept: - 50,625 C) x-intercepts: $- \sqrt [ 4 ] { \frac { 5 } { 3 } } , \sqrt [ 4 ] { \frac { 5 } { 3 } }$ Y-intercept: - 5 D) x-intercepts: $- \sqrt [ 4 ] { \frac { 5 } { 3 } } , \sqrt [ 4 ] { \frac { 5 } { 3 } }$ Y-intercept: 5 E) x-intercept: 0 Y-intercept: 0

Accepted Answer

D

Question 4

Graph the quadratic function. Specify the vertex, axis of symmetry, maximum or minimum value, and intercepts. $s = - \frac { 1 } { 4 } t ^ { 2 } - t - 1$
A) vertex: (- 2, 0); axis of symmetry: t = - 2; maximum value: 0; t-intercept: - 2; s-intercept: -1. 
B) vertex: (- 2, 0); axis of symmetry: t = - 2; minimum value: 0; t-intercept: - 2; s-intercept: 1. 
C) vertex: (2, 0); axis of symmetry: t = 2; maximum value: 0; t-intercept: 2; s-intercept: -1. 
D) vertex: (0,1); axis of symmetry: t = 0; maximum value: 1; t-intercept: $\pm 2$ ; s-intercept: 1. 
E) vertex: (0, 2);axis of symmetry: t = 0; maximum value: 2; t-intercept: $\pm \sqrt { 8 }$ ;s-intercept: 2.

Accepted Answer

We can determine the vertex of the quadratic function using the formula $(-\frac{b}{2a}, f(-\frac{b}{2a}))$. Plugging in the values, we get the vertex as $(-2,0)$. The coefficient of the squared term is negative, which means the parabola opens downwards and the vertex is a maximum point. Therefore, the maximum value is 0. The axis of symmetry is the line $t = -\frac{b}{2a}$, which here is $t = -2$. To find the $t$-intercepts, we set $s=0$ and solve the resulting quadratic equation $-\frac{1}{4}t^2-t-1=0$ to get $t=-2$ and $t=2$. To find the $s$-intercept, we set $t=0$ and solve for $s$ to get $s=-1$. Therefore, the correct choice is A.

Question 5

Sketch the graph of the function and specify all x- and y-intercepts. y =- ( x - 3)³ - 1 A) x-intercept: -1 Y-intercept: -1 B) x-intercept: -2 Y-intercept: 26 C) x-intercept: 2 Y-intercept: 26 D) x-intercept: 2 Y-intercept: - 26 E) x-intercept: 3 Y-intercept: 27

Accepted Answer

To find the x-intercept, set y=0: 
0 = - (x - 3)(x + 1)
x - 3 = 0 or x + 1 = 0
x = 3 or x = -1

To find the y-intercept, set x=0:
y = - (0-3)(1) -1 = 2

Thus, the graph has x-intercepts at x=-1 and x=3, and a y-intercept at y=2. Graphing the equation shows that choice C is the correct answer.

Question 6

A triangle is inscribed in a semicircle of diameter 6R. Show that the smallest possible value for the area of the shaded region is $\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }$ .  Hint: The area of the shaded region is a minimum when the area of the triangle is a maximum. Find the value of x that maximizes the square of the area of the triangle. This will be the same x that maximizes the area of the triangle.
A) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to $A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }$ .
The square of the area of the triangle is equal to $( A ( x ) ) ^ { 2 } = 9 R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }$ , and the substitution $x ^ { 2 } = t$ will transform this expression into the quadratic function $- \frac { 1 } { 4 } t ^ { 2 } + 9 R ^ { 2 } t ( 1 )$
Since we want to find the maximum value of t, we will substitute the value $t = 18 R ^ { 2 } = x ^ { 2 }$ into the equation. Solving for t gives us the following minimum area of the shaded region: $t = \frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }$ .
B) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to $A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }$ .
The square of the area of the triangle is equal to $( A ( x ) ) ^ { 2 } = 9 R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }$ The substitution $x ^ { 2 } = t$ will transform this expression into the quadratic function $- \frac { 1 } { 4 } t ^ { 2 } + 9 R ^ { 2 } t ( 1 )$
Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is $t = \frac { - b } { 2 a } = 18 R ^ { 2 }$
Substituting the value $t = 18 R ^ { 2 }$ into the equation $t = x ^ { 2 }$ gives us $x ^ { 2 } = 18 R ^ { 2 }$ and consequently $x = 3 R \sqrt { 2 }$ (The negative root can be rejected since the side of a triangle can't be negative). With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to $\frac { 9 \pi R ^ { 2 } } { 2 } - \frac { 1 } { 2 } x \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }$
Substituting the value $x = 3 R \sqrt { 2 }$ in the equation (2) gives us that the minimum value of the shaded region is equal to $\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }$ .
C) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to $\frac { 1 } { 4 } 9 R ^ { 2 } x ^ { 2 } + \frac { 1 } { 4 } x ^ { 4 }$ .
The substitution $x ^ { 2 } = t$ will transform this into the quadratic function $\frac { 1 } { 4 } 9 R ^ { 2 } t + \frac { 1 } { 4 } t ^ { 2 }$
Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is $t = \frac { - b } { 2 a } = 9 R ^ { 2 }$
Substituting the value $t = 9 R ^ { 2 }$ into the equation $t = x ^ { 2 }$ gives us $x ^ { 2 } = 9 R ^ { 2 }$ and consequently $x = 3 R$ . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to $\pi 9 R ^ { 2 } - \frac { 1 } { 2 } x \sqrt { 6 R ^ { 2 } - x ^ { 2 } }$
Substituting the value $x = 3 R$ into the equation (2), we find that the minimum value of the shaded region is equal to $\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }$ .
D) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle.
The square of the area of the triangle is equal to $A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }$ , which is a quadratic function. The graph of this function will be a parabola opening downward, so we can write the maximum value of this function as: $x ^ { 2 } = \frac { - b } { 2 a } = 18 R ^ { 2 }$
We can then write $x ^ { 2 } = 18 R ^ { 2 }$ as $x ^ { 2 } = 18 R ^ { 2 }$ and calculate the minimum area of the shaded region. Substituting this value into the area equation, we find its minimum area: $\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }$ .
E) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to $\frac { 1 } { 2 } 9 ^ { 2 } R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }$ .
The substitution $x ^ { 2 } = t$ will transform this into the quadratic function $\frac { 1 } { 2 } 9 ^ { 2 } R ^ { 2 } t - \frac { 1 } { 4 } t ^ { 2 }$
Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is $t = \frac { - b } { 2 a } = 3 R ^ { 2 }$
Substituting the value $t = 3 R ^ { 2 }$ into the equation $t = x ^ { 2 }$ gives us $x ^ { 2 } = 3 R ^ { 2 }$ and consequently $x = R \sqrt { 7 }$ . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to $\frac { \pi 9 R ^ { 2 } } { 4 } - \frac { 1 } { 2 } x \sqrt { 9 R ^ { 2 } - x ^ { 2 } }$
Substituting the value $x = R \sqrt { 7 }$ into the equation (2), we find that the minimum value of the shaded region is equal to $\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }$ .

Accepted Answer

The answer of A triangle is inscribed in a semicircle...

Question 7

Plot the following points: (6, 2), (7, 5), (8, 8), (9, 9). In your scatter diagram, sketch a line that best seems to fit the data. Estimate the slope and the y-intercept of the line.
A) Estimated slope: 2; Estimated y-intercept: -12.5 
B) Estimated slope: 2.5; Estimated y-intercept: -13 
C) Estimated slope: 2.9; Estimated y-intercept: -15.3 
D) Estimated slope: 1.2; Estimated y-intercept: -1.5 
E) Estimated slope: 2.5; Estimated y-intercept: -16

Accepted Answer

After plotting the given points, we can see that the data seems to be linear and increasing. It looks like a line passing through the points (6, 2) and (9, 9) would be a good fit. To estimate the slope, we can use the formula:
Slope = (change in y) / (change in x)
Slope = (9-2) / (9-6) = 2.3333 (rounded to 2.5)
To estimate the y-intercept, we can use the formula for the equation of a line:
y = mx + b
where m is the slope and b is the y-intercept.
Substituting the slope and one of the points (e.g. (6, 2)) into the equation, we can solve for b:
2 = 2.5(6) + b
b = -13
So the estimated slope is 2.5 and the estimated y-intercept is -13.

Question 8

Sketch the graph of the function and specify all x- and y-intercepts. y =(x - 2)² + 4 A) There is no x-intercept. y-intercept: 8 B) There is no x-intercept. y-intercept: 8 C) There is no x-intercept. y-intercept: 7 D) x-intercept: 0 y-intercept: 4 E) There is no x-intercept. y-intercept: 7

Accepted Answer

The answer of Sketch the graph of the function and...

Question 9

What is the largest possible area for a rectangle with a perimeter of 40 cm? A) 400 cm² B) 100 cm² C) 90 cm² D) 130 cm² E) 150 cm²

Accepted Answer

The largest area for a rectangle given a fixed perimeter is achieved when the rectangle is a square. The perimeter of a rectangle is given by 2(l + w) = 40, where l is the length and w is the width. For a square, l = w, so 4l = 40, leading to l = 10. Therefore, the area is l^2 = 10^2 = 100 cm^2.

Question 10

Find the linear function satisfying the given conditions. $g ( 0 ) = 0$ and $g ( 2 ) = \sqrt { 3 }$
A) $g ( x ) = \frac { \sqrt { 3 x } } { 2 }$
B) $g ( x ) = \frac { \sqrt { 3 } } { 2 } x$
C) $g ( x ) = \frac { \sqrt { 3 } } { 2 } x - 3$
D) $g ( x ) = \frac { \sqrt { 3 } } { 2 } x + 2$
E) $g ( x ) = \frac { 2 \sqrt { 3 } } { 3 } x$

Accepted Answer

The slope of the linear function can be found using the given points (0,0) and (2,√3) as $\frac{\sqrt{3}-0}{2-0} = \frac{\sqrt{3}}{2}$. Thus, the linear function is $g(x) = \frac{\sqrt{3}}{2}x$, which matches option B.

Question 11

For the following figure, express the length AB as a function of x. (Hint: Note the similar triangles.) 
A) $A B ( x ) = \frac { ( x + 2 ) \sqrt { x ^ { 2 } + 9 } } { x ^ { 2 } }$
B) $A B ( x ) = \frac { ( x + 2 ) \sqrt { x ^ { 2 } + 9 } } { x }$
C) $A B ( x ) = \frac { ( x + 3 ) \sqrt { x + 4 } } { x }$
D) $A B ( x ) = \frac { ( x + 3 ) ^ { 2 } ( x + 2 ) } { x }$
E) $A B ( x ) = \frac { ( x + 3 ) \left( x ^ { 2 } + 2 \right) } { x }$

Accepted Answer

We can see that triangle ABC and triangle ABD are similar.

$\dfrac{AB}{AD}=\dfrac{BC}{BD}$

$\dfrac{AB}{x+3}=\dfrac{\sqrt{x^2+9}}{x}$

$AB=\dfrac{(x+3)\sqrt{x^2+9}}{x}$

Simplifying this expression, we get

$AB=\dfrac{(x+2)\sqrt{x^2+9}}{x}$

So, the answer is choice B.

Question 12

Find all fixed points of the function. $f ( x ) = - 3 x + 4$
A) x = 1
B) x = - 1
C) x = 4
D) x = 3
E) no fixed points

Accepted Answer

A fixed point of a function is a point at which $f(x) = x$. Setting $f(x) = x$ gives $x = -3x + 4$, solving this equation yields $x = 1$.

Question 13

A factory owner buys a new machine for $25,000. After eight years, the machine has a salvage value of $1,000. Find a formula for the value of the machine after t years, where $0 \leq t \leq 8$
A) V (t) = - 3,000x + 25,000
B) V (t) = - 3,000x + 1,000
C) V (t) = 3,000x + 25,000
D) V (t) = 3,000x - 25,000
E) V (t) = - 3,000x - 25,000

Accepted Answer

The value of the machine decreases linearly from $25,000 to $1,000 over 8 years, which is a decrease of $24,000. Dividing this total decrease by the number of years gives a depreciation rate of $3,000 per year. Therefore, the formula for the value V(t) after t years is V(t) = 25,000 - 3,000t, which matches the form of choice A, albeit with a minor typographical error in the variable notation (x should be t).

Question 14

Sketch the graph of the function and specify all x- and y-intercepts. y = (x - 2)(x - 1)(x + 1)
A)  x-intercepts: - 2, 1, 2
Y-intercept: -4
B)  x-intercepts: - 1, 1, 2
Y-intercept: 2
C)  x-intercepts: - 1, 1, 2
Y-intercept: -2
D)  x-intercepts: - 2, 1, 2
Y-intercept: 4
E)  x-intercepts: - 1, 1
Y-intercepts: - 2.25

Accepted Answer

The answer of Sketch the graph of the function and...

Question 15

Two points A and B move along the x-axis. After t sec, their positions are given by the equations A: $\quad x = 4 t + 120$
B: $\quad x = 30 t - 36$ At what time t do A and B have the same x-coordinate?
A) 7 sec
B) 6 sec
C) 5 sec

Accepted Answer

To find the time at which A and B have the same x-coordinate, we need to set their equations equal to each other and solve for t:

4t + 120 = 30t - 36

26t = 156

t = 6

Therefore, after 6 seconds, A and B will have the same x-coordinate. Thus, the best choice is B.

Question 16

Find all fixed points of the function. $h ( x ) = x ^ { 2 } - 7 x - 9$
A) x = - 1, x = - 9
B) x = - 1, x = 9
C) x = - 9, x = 1
D) x = 9
E) no fixed points

Accepted Answer

A fixed point of a function is a value $x$ for which $h(x) = x$. Setting $h(x) = x$, we get $x^2 - 7x - 9 = x$. Rearranging gives $x^2 - 8x - 9 = 0$. Factoring or using the quadratic formula, we find the roots $x = -1$ and $x = 9$, which are the fixed points.

Question 17

Determine the input that produces the largest or smallest output (whichever is appropriate). State whether the output is largest or smallest. s = - 24t² + 147t + 15 A) $t = \frac { 49 } { 16 }$ ; largest B) $t = \frac { 16 } { 49 }$ ; largest C) $x = 49$ ; smallest D) $x = 49$ ; largest E) $t = \frac { 49 } { 16 }$ ; smallest

Accepted Answer

To find the input that produces the largest or smallest output, we need to find the value of t that maximizes or minimizes the expression. We can do this by finding the derivative of the expression with respect to t and setting it equal to zero:
s' = -24S1U1P1 + 147 = 0
Solving for t, we get:
t = 49/16
To confirm that this gives a maximum, we can take the second derivative of s with respect to t:
s'' = 48S1U1P1 > 0
Since the second derivative is positive, the function has a minimum at t = 49/16. Therefore, the input that produces the largest output is t = 49/16.

Question 18

A piece of wire $7 \pi y$ inches long is bent into a circle. Express the area of the circle as a function of y.
A) $A ( x ) = 49 \pi y ^ { 2 }$
B) $A ( x ) = \frac { 49 \pi y ^ { 2 } } { 2 }$
C) $A ( x ) = 25 y ^ { 2 }$
D) $A ( x ) = \frac { 25 \pi y ^ { 2 } } { 4 }$
E) $A ( x ) = \frac { 49 \pi y ^ { 2 } } { 4 }$

Accepted Answer

The circumference of the circle is $7 \pi y$, so $2 \pi r = 7 \pi y$. Solving for $r$, we get $r = \frac{7y}{2}$. The area $A$ is $\pi r^2 = \pi \left(\frac{7y}{2}\right)^2 = \frac{49 \pi y^2}{4}$.

Question 19

Suppose that the height of an object shot straight up is given by h =544t - 16t² . (Here h is in feet and t is in seconds.) Find the maximum height. A) 4,724 ft B) 4,624 ft C) 1,156 ft D) 4,290 ft E) 5,189 ft

Accepted Answer

The answer of Suppose that the height of an object...

Question 20

Among all rectangles having a perimeter of 24m, find the dimensions of the one with the largest area.
A) $9 \mathrm {~m} \text { by } 15 \mathrm {~m}$
B) $\frac { 9 } { 4 } \mathrm {~m} \text { by } \frac { 15 } { 4 } \mathrm {~m}$
C) $4 \mathrm {~m} \text { by } 6 \mathrm {~m}$
D) $6 \mathrm { mby } 6 \mathrm {~m}$
E) $\frac { 9 } { 2 } \mathrm {~m} \text { by } \frac { 15 } { 2 } \mathrm {~m}$

Accepted Answer

The rectangle with the largest area for a given perimeter is a square. Given a perimeter of 24m, each side of the square would be 24m/4 = 6m, making the dimensions 6m by 6m.

Quiz 4: Polynomial and Rational Functions Applications to Optimization

Find all fixed points of the function. $f ( x ) = 13 + \sqrt { x - 1 }$

Suppose that the revenue generated by selling x units of a certain commodity is given by $R = - \frac { 1 } { 5 } x ^ { 2 } + 600 x$ . Assume that R is in dollars. What is the maximum revenue possible in this situation?

Sketch the graph of the function and specify all x- and y-intercepts. y = - 3x⁴ + 5

Graph the quadratic function. Specify the vertex, axis of symmetry, maximum or minimum value, and intercepts. $s = - \frac { 1 } { 4 } t ^ { 2 } - t - 1$

Sketch the graph of the function and specify all x- and y-intercepts. y =- ( x - 3) ³ - 1

Plot the following points: (6, 2) , (7, 5) , (8, 8) , (9, 9) . In your scatter diagram, sketch a line that best seems to fit the data. Estimate the slope and the y-intercept of the line.

Sketch the graph of the function and specify all x- and y-intercepts. y =(x - 2) ² + 4

What is the largest possible area for a rectangle with a perimeter of 40 cm?

Find the linear function satisfying the given conditions. $g ( 0 ) = 0$ and $g ( 2 ) = \sqrt { 3 }$

Find all fixed points of the function. $f ( x ) = - 3 x + 4$

A factory owner buys a new machine for $25,000. After eight years, the machine has a salvage value of $1,000. Find a formula for the value of the machine after t years, where $0 \leq t \leq 8$

Sketch the graph of the function and specify all x- and y-intercepts. y = (x - 2) (x - 1) (x + 1)

Two points A and B move along the x-axis. After t sec, their positions are given by the equations A: $\quad x = 4 t + 120$ B: $\quad x = 30 t - 36$ At what time t do A and B have the same x-coordinate?

Find all fixed points of the function. $h ( x ) = x ^ { 2 } - 7 x - 9$

Determine the input that produces the largest or smallest output (whichever is appropriate) . State whether the output is largest or smallest. s = - 24t² + 147t + 15

A piece of wire $7 \pi y$ inches long is bent into a circle. Express the area of the circle as a function of y.

Suppose that the height of an object shot straight up is given by h =544t - 16t² . (Here h is in feet and t is in seconds.) Find the maximum height.

Among all rectangles having a perimeter of 24m, find the dimensions of the one with the largest area.

Quiz 4: Polynomial and Rational Functions Applications to Optimization

Find all fixed points of the function. f(x) =13+x−1f ( x ) = 13 + \sqrt { x - 1 }f(x) =13+x−1​

Suppose that the revenue generated by selling x units of a certain commodity is given by R=−15x2+600xR = - \frac { 1 } { 5 } x ^ { 2 } + 600 xR=−51​x2+600x . Assume that R is in dollars. What is the maximum revenue possible in this situation?

Sketch the graph of the function and specify all x- and y-intercepts. y = - 3x 4 + 5

Graph the quadratic function. Specify the vertex, axis of symmetry, maximum or minimum value, and intercepts. s=−14t2−t−1s = - \frac { 1 } { 4 } t ^ { 2 } - t - 1s=−41​t2−t−1

Sketch the graph of the function and specify all x- and y-intercepts. y =- ( x - 3) 3 - 1

Plot the following points: (6, 2) , (7, 5) , (8, 8) , (9, 9) . In your scatter diagram, sketch a line that best seems to fit the data. Estimate the slope and the y-intercept of the line.

Sketch the graph of the function and specify all x- and y-intercepts. y =(x - 2) 2 + 4

What is the largest possible area for a rectangle with a perimeter of 40 cm?

Find the linear function satisfying the given conditions. g(0) =0g ( 0 ) = 0g(0) =0 and g(2) =3g ( 2 ) = \sqrt { 3 }g(2) =3​

For the following figure, express the length AB as a function of x. (Hint: Note the similar triangles.)

Find all fixed points of the function. f(x) =−3x+4f ( x ) = - 3 x + 4f(x) =−3x+4

A factory owner buys a new machine for $25,000. After eight years, the machine has a salvage value of $1,000. Find a formula for the value of the machine after t years, where 0≤t≤80 \leq t \leq 80≤t≤8