alg 2

Function Definitions

Functions are defined as follows: $f(a) = 2a - 3$ and $g(a) = -2a + 3$. These definitions serve as the basis for various operations.

Function Addition

To find $(f + g)(a)$, we sum the definitions of $f(a)$ and $g(a)$. This involves calculating $(2a - 3) + (-2a + 3)$. Upon simplification, the result is$= 2a - 3 - 2a + 3 = 0$. Therefore, the sum of these two specific functions is 0.

Solving Linear Equations

A linear equation such as $2x - 3 = 22 + 3$ can be solved by first simplifying the right side to $2x - 3 = 25$. Next, add 3 to both sides, which yields $2x = 28$. Finally, dividing by 2 gives $x = 14$, which is the solution to the equation.

Function Evaluation

For $f(x) = 2x$, evaluating the function at 6 results in $f(6) = 2(6) = 12$. Similarly, for $g(x) = 3x - 4$, evaluating at 6 gives $g(6) = 3(6) - 4 = 18 - 4 = 14$.

Function Subtraction

Using the previously evaluated values, to find $(f-g)(6)$, we subtract $g(6)$ from $f(6)$. This calculation is $12 - 14 = -2$. Thus, the difference of the functions at $x=6$ is -2.

Additional Functions

Further functions introduced include $h(a) = 3a + 3$ and a new definition for $g(a) = a^3 - 5a$. These functions are used for compositions.

Composition of Functions

To compute $(h \text{ o } g)(-4)$, we first evaluate the inner function $g(-4)$. Substituting -4 into $g(a)$ gives $(-4)^3 - 5(-4) = -64 + 20 = -44$. Next, we evaluate $h$ at this result: $h(-44) = 3(-44) + 3 = -132 + 3 = -129$. Hence, $(h \text{ o } g)(-4) = -129$.

Further Function Evaluation

Consider two functions, $h(x) = 4x + 3$ and $g(x) = 4x + 1$. To find $(h + g)(-9)$, we evaluate each function at -9 and add the results: $(4(-9) + 3) + (4(-9) + 1)$. This simplifies to $(-36 + 3) + (-36 + 1) = -33 + -35 = -68$.

Inverse of Functions

To find the inverse of a function, such as $h(x) = -2x - 5$, the following steps are taken: First, replace $h(x)$ with $y$, yielding $y = -2x - 5$. Next, solve for $x$ in terms of $y$. Adding 5 to both sides gives $y + 5 = -2x$. Dividing by -2 results in $x = -\frac{(y + 5)}{2}$. Therefore, the inverse function is $h^{-1}(y) = -\frac{(y+5)}{2}$.

Systems of Equations

Solve Each System

1. Given the system $y = 2x - 12$ and $y = -8x + 8$, we can set the two expressions for $y$ equal to each other: $2x - 12 = -8x + 8$. Rearranging terms, we get $10x = 20$, which means $x = 2$. Substituting $x = 2$ back into the first equation yields $y = 2(2) - 12 = 4 - 12 = -8$. The solution to this system is $(2, -8)$.

2. For the system $5x + y = 21$ and $4x - y = -12$, adding both equations eliminates $y$: $(5x + y) + (4x - y) = 21 + (-12)$. This simplifies to $9x = 9$, so $x = 1$. Substituting $x = 1$ into the first equation gives $5(1) + y = 21$, which means $5 + y = 21$. Solving for $y$ gives $y = 16$. The solution to this system is $(1, 16)$.

More Systems of Equations

Another system of equations provided is $-5x - 7x + 9y = -27$, which appears to be incomplete or a single linear equation in two variables rather than a system.

More Functions and Solving Complex Equations

To solve the quadratic equation $9n^2 + 5 = -79$, we first rearrange it to $9n^2 = -84$. One must check for extraneous solutions. Additional quadratic equations to solve include $a^2 - 14a - 46 = 5$, $x^2 = -5 + 6x$, $n^2 - 12 = -2n$, $n^2 - n + 7 = 7$, and $5m^2 - 8 = -82$, each requiring appropriate methods for quadratic solutions.

Graphing Functions

Various functions are intended for graphing, including $y = -x^2$, $y = -(x - 2)^2 + 2$, $y = x^2 + 2x - 3$, and $y = 2x^2 + 4x + 1$. These represent a parabola opening downwards, a parabola shifted, and two general quadratic functions, respectively.

Using the Discriminant

The discriminant can be used to determine the number of solutions for quadratic equations. For $8x^2 + 8x + 5 = 0$ and $-4v^2 - 2v - 3 = 0$, calculating $b^2 - 4ac$ will reveal whether there are real, one real, or no real solutions.

Absolute Value Graphs

Examples of absolute value functions to graph include $y = |x| - 2$, $y = |x - 2| - 3$, and $y = -|x| + 2$. These equations typically form V-shaped graphs, which can be shifted or inverted.

Function Definitions Functions are defined as follows: $f(a) = 2a - 3$ and $g(a) = -2a + 3$. These definitions serve as the basis for various operations. #### Function Addition To find $(f + g)(a)$, we sum the definitions of $f(a)$ and $g(a)$. This involves calculating $(2a - 3) + (-2a + 3)$. Upon simplification, the result is$= 2a - 3 - 2a + 3 = 0$. Therefore, the sum of these two specific functions is 0. #### Solving Linear Equations A linear equation such as $2x - 3 = 22 + 3$ can be solved by first simplifying the right side to $2x - 3 = 25$. Next, add 3 to both sides, which yields $2x = 28$. Finally, dividing by 2 gives $x = 14$, which is the solution to the equation. #### Function Evaluation For $f(x) = 2x$, evaluating the function at 6 results in $f(6) = 2(6) = 12$. Similarly, for $g(x) = 3x - 4$, evaluating at 6 gives $g(6) = 3(6) - 4 = 18 - 4 = 14$. #### Function Subtraction Using the previously evaluated values, to find $(f-g)(6)$, we subtract $g(6)$ from $f(6)$. This calculation is $12 - 14 = -2$. Thus, the difference of the functions at $x=6$ is -2. #### Additional Functions Further functions introduced include $h(a) = 3a + 3$ and a new definition for $g(a) = a^3 - 5a$. These functions are used for compositions. #### Composition of Functions To compute $(h \text{ o } g)(-4)$, we first evaluate the inner function $g(-4)$. Substituting -4 into $g(a)$ gives $(-4)^3 - 5(-4) = -64 + 20 = -44$. Next, we evaluate $h$ at this result: $h(-44) = 3(-44) + 3 = -132 + 3 = -129$. Hence, $(h \text{ o } g)(-4) = -129$. #### Further Function Evaluation Consider two functions, $h(x) = 4x + 3$ and $g(x) = 4x + 1$. To find $(h + g)(-9)$, we evaluate each function at -9 and add the results: $(4(-9) + 3) + (4(-9) + 1)$. This simplifies to $(-36 + 3) + (-36 + 1) = -33 + -35 = -68$. #### Inverse of Functions To find the inverse of a function, such as $h(x) = -2x - 5$, the following steps are taken: First, replace $h(x)$ with $y$, yielding $y = -2x - 5$. Next, solve for $x$ in terms of $y$. Adding 5 to both sides gives $y + 5 = -2x$. Dividing by -2 results in $x = -\frac{(y + 5)}{2}$. Therefore, the inverse function is $h^{-1}(y) = -\frac{(y+5)}{2}$. #### Systems of Equations ##### Solve Each System 1. Given the system $y = 2x - 12$ and $y = -8x + 8$, we can set the two expressions for $y$ equal to each other: $2x - 12 = -8x + 8$. Rearranging terms, we get $10x = 20$, which means $x = 2$. Substituting $x = 2$ back into the first equation yields $y = 2(2) - 12 = 4 - 12 = -8$. The solution to this system is $(2, -8)$. 2. For the system $5x + y = 21$ and $4x - y = -12$, adding both equations eliminates $y$: $(5x + y) + (4x - y) = 21 + (-12)$. This simplifies to $9x = 9$, so $x = 1$. Substituting $x = 1$ into the first equation gives $5(1) + y = 21$, which means $5 + y = 21$. Solving for $y$ gives $y = 16$. The solution to this system is $(1, 16)$. #### More Functions and Solving Complex Equations To solve various quadratic equations: 1. For $9n^2 + 5 = -79$: Subtract 5 from both sides to get $9n^2 = -84$. Divide by 9: $n^2 = -\frac{84}{9} = -\frac{28}{3}$. Since the square of a real number cannot be negative, there are no real solutions. (Complex solutions: $n = \pm i\sqrt{\frac{28}{3}}$) 2. For $a^2 - 14a - 46 = 5$: Subtract 5 from both sides to get $a^2 - 14a - 51 = 0$. Use the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1, b=-14, c=-51$. $a = \frac{14 \pm \sqrt{(-14)^2 - 4(1)(-51)}}{2(1)} = \frac{14 \pm \sqrt{196 + 204}}{2} = \frac{14 \pm \sqrt{400}}{2} = \frac{14 \pm 20}{2}$. Solutions are $a<em>1 = 17$ and $a</em>2 = -3$. 3. For $x^2 = -5 + 6x$: Rearrange to standard quadratic form $x^2 - 6x + 5 = 0$. Factor the quadratic: $(x - 1)(x - 5) = 0$. Solutions are $x = 1$ and $x = 5$. 4. For $n^2 - 12 = -2n$: Rearrange to standard quadratic form $n^2 + 2n - 12 = 0$. Use the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1, b=2, c=-12$. $n = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-12)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 48}}{2} = \frac{-2 \pm \sqrt{52}}{2} = \frac{-2 \pm 2\sqrt{13}}{2}$. Solutions are $n = -1 \pm \sqrt{13}$. 5. For $n^2 - n + 7 = 7$: Subtract 7 from both sides to get $n^2 - n = 0$. Factor out $n$: $n(n - 1) = 0$. Solutions are $n = 0$ and $n = 1$. 6. For $5m^2 - 8 = -82$: Add 8 to both sides to get $5m^2 = -74$. Divide by 5: $m^2 = -\frac{74}{5}$. Since the square of a real number cannot be negative, there are no real solutions. (Complex solutions: $m = \pm i\sqrt{\frac{74}{5}}$) #### Graphing Functions Examples of graphing quadratic and absolute value functions: 1. $y = -x^2$: This is a parabola opening downwards with its vertex at $(0,0)$. To graph, plot the vertex. Choose symmetric $x$ values (e.g., $x = \pm 1, y = -1$; $x = \pm 2, y = -4$) and plot the corresponding points to sketch the curve. 2. $y = -(x - 2)^2 + 2$: This is a parabola opening downwards, with its vertex shifted to $(2,2)$. To graph, plot the vertex $(2,2)$. Then, similar to $y = -x^2$, move 1 unit horizontally from the vertex and 1 unit down (e.g., to $(1,1)$ and $(3,1)$). Move 2 units horizontally and 4 units down (e.g., to $(0,-2)$ and $(4,-2)$). 3. $y = x^2 + 2x - 3$: This is a parabola opening upwards. To graph, first find the vertex using $x = -\frac{b}{2a}$. For this function, $x = -\frac{2}{2(1)} = -1$. Substitute $x=-1$ into the equation to find $y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$. So the vertex is $(-1,-4)$. Find x-intercepts by setting $y=0$ ($(x+3)(x-1)=0 \Rightarrow x=-3, x=1$) and the y-intercept by setting $x=0$ ($y=-3$). Plot these points and sketch the parabola. 4. $y = 2x^2 + 4x + 1$: This is a parabola opening upwards, narrower than $y = x^2$. To graph, find the vertex using $x = -\frac{b}{2a}$. For this function, $x = -\frac{4}{2(2)} = -1$. Substitute $x=-1$ into the equation to find $y = 2(-1)^2 + 4(-1) + 1 = 2 - 4 + 1 = -1$. So the vertex is $(-1,-1)$. Find the y-intercept by setting $x=0$ ($y=1$). Plot additional points by choosing $x$ values and calculating $y$. #### Using the Discriminant The discriminant ($D = b^2 - 4ac$) can be used to determine the number of real solutions for quadratic equations of the form $ax^2 + bx + c = 0$. * If $D > 0$, there are two distinct real solutions. * If $D = 0$, there is exactly one real solution. * If $D < 0$, there are no real solutions (two complex solutions). 1. For $8x^2 + 8x + 5 = 0$: Here, $a=8, b=8, c=5$. Calculate the discriminant: $D = (8)^2 - 4(8)(5) = 64 - 160 = -96$. Since $D < 0$, there are no real solutions. 2. For $-4v^2 - 2v - 3 = 0$: Here, $a=-4, b=-2, c=-3$. Calculate the discriminant: $D = (-2)^2 - 4(-4)(-3) = 4 - 48 = -44$. Since $D < 0$, there are no real solutions. #### Absolute Value Graphs Examples of absolute value functions, which typically form V-shaped graphs: 1. $y = |x| - 2$: This is a V-shaped graph opening upwards, with its vertex at $(0,-2)$. To graph, plot the vertex. From the vertex, for every 1 unit moved horizontally, move 1 unit vertically upwards (due to the coefficient of $x$ being 1). For example, points include $(1,-1)$ and $(-1,-1)$. 2. $y = |x - 2| - 3$: This is a V-shaped graph opening upwards, with its vertex shifted to $(2,-3)$. To graph, plot the vertex $(2,-3)$. From this vertex, move 1 unit horizontally and 1 unit vertically upwards (e.g., to $(3,-2)$ and $(1,-2)$). 3. $y = -|x| + 2$: This is an inverted V-shaped graph opening downwards, with its vertex at $(0,2)$. To graph, plot the vertex $(0,2)$. From the vertex, for every 1 unit moved horizontally, move 1 unit vertically downwards (due to the negative sign before $|x|$). For example, points include $(1,1)$ and $(-1,1)$.