#1**+2 **

The range is the set of all possible outputs that a function can produce.

\(f(x)=4|x+5|\)

The absolute value of any number will can result in a positive number or 0, and 4 multiplied by a positive number or 0 does not change this property at all. Therefore, the range is the following:

\(\text{Range}:\{\mathbb{R}|\hspace{1mm}y\geq0\}\)

This means that the range can be all nonnegative numbers. In interval notation, it would look like the following:

\(\text{Range}:[0,+\infty)\)

How do we find the intercepts? To do it without a graph, you can figure out by setting x=0 and y=0 and solving in each case. Let's do that.

\(y=4|x+5|\) | Plug in 0 for x. This time, we are solving for the y-intercept. |

\(y=4|0+5|\) | Simplify inside the absolute value first. |

\(y=4|5|\) | |

\(y=4*5=20\) | We have now determined the coordinates of the y-intercept. |

\((0,20)\) | This is the exact coordinates of the y-intercept. |

Let's do the exact same process. This time, however, we set y=0 to find the x-intecept.

\(y=4|x+5|\) | Set y equal to 0. |

\(0=4|x+5|\) | Divide by 4 on both sides. |

\(0=|x+5|\) | The absolute value always splits an equation into its plus or minus. However, 0 is neither positive nor negative, so there aren't 2 equations that one can set up. |

\(x+5=0\) | |

\(x=-5\) | We have now determined the x-intercept, as well. |

\((-5,0)\) | |

TheXSquaredFactor
Sep 17, 2017

#1**+1 **

A constant term is the term that is unchanging. 6, for example, is a constant. 12.5 is a constant; they don't change.

To figure out \(f(g(x))\), let's break this down bit by bit. Since we already know that \(g(x)=x^3+5x^2+9x-2\), this means that \(f(g(x))=f(x^3+5x^2+9x-2)\).

\(f(x)=x^3-6x^2+3x-4\) | Now, substitute f(g(x)) into all instances of x. |

\(f(g(x))=(x^3+5x^2+9x-2)^3-6(x^3+5x^2+9x-2)^2+3(x^3+5x^2+9x-2)-4\) | Luckily, however, we only care about the constant terms. Let's deal with one term at a time. |

\((x^3+5x^2+9x-2)^3\) | We only care about the constant term, so do -2^3=-8 |

\((-2)^3=-8\) | Let's worry about the second term. |

\(-6(x^3+5x^2+9x-2)^2\) | Let's do the exact same process. |

\(-6*(-2)^2=-6*4=-24\) | And of course, the next term, as well. |

\(3(x^3+5x^2+9x-2)\) | |

\(3*-2=-6\) | And the final term, which happens to be a constant. |

\(-4=-4\) | Now, add all of those together to get the constant term. |

\(-8-24-6-4 = -42\) | This is value of the constant term. |

TheXSquaredFactor
Sep 17, 2017

#6**+1 **

BOSEOK is correct; the equation is x=3.

Contrary to what Gh0sty stated, the slope is defined when a linear function happens to be horizontal. Let's figure out the slope of the graph provided by Gh0sty. I will use the points \((3,-2)\) and \((0,-2)\). Let's find the slope.

Of course, the formula for slope is the following:

\(m=\frac{y_2-y_1}{x_2-x_1}\)

\(m=\frac{-2-(-2)}{0-3}\) | Continue to simplify the fraction. |

\(m=\frac{0}{-3}=0\) | |

The slope is 0. 0 is a valid number for the slope. If you type into Demos y=0x-2, the line will appear exactly as the picture above.

Let's try calculating the slope of the line x=-3. Two arbitrary points on the line are \((-3,2)\) and \((-3,0)\)

What is the slope of this? Let's try using the formula again. Let's see what happens.

\(\frac{0-2}{-3-(-3)}\) | Simplify both the numerator and the denominator. |

\(\frac{-2}{0}\) | |

Of course, any number divided by 0 is undefined and therefore it has an undefined slope.

Therefore, BOSEOK's answer is correct because it meets both criteria of

1) A line with an undefined slope

2) A line that passes through the point \((3,-2)\)

And therefore, \(x=3\) is correct.

TheXSquaredFactor
Sep 17, 2017

#2**+1 **

I'll show my method of proving. Of course, the standard form of a quadratic equation is in the following form of \(ax^2+bx+c=0\). If I start with the quadratic formula \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\), then all I should have to do is utilize some algebraic manipulation to arrive to the standard form. Let's do that!

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) | First, let's eliminate the fraction on the right hand side by multiplying by 2a on both sides. |

\(2ax=-b\pm\sqrt{b^2-4ac}\) | Add b to both sides of the equation. |

\(2ax+b=\pm\sqrt{b^2-4ac}\) | Square both sides. Normally, you would have to consider both cases, but squaring always results in the positive answer--no matter if the original number is positive or negative. Therefore, both will result to the same thing. |

\((2ax+b)^2=b^2-4ac\) | Expand the left hand side of the equation by knowing that \((x+y)^2=x^2+2xy+y^2\). |

\((2ax)^2+2(2ax)(b)+b^2=b^2-4ac\) | Subtract b^2 from both sides. |

\((2ax)^2+2(2ax)(b)=-4ac\) | Simplify the left hand side by dealing with the multiplication and the exponent. |

\(4a^2x^2+4axb=-4ac\) | Divide by the GCF of both sides, which is 4a. |

\(ax^2+bx=-c\) | And finally, subtract c on both sides. |

\(ax^2+bx+c=0\) | |

I have now proven that the quadratic formula works for all numbers for A,B, and C.

TheXSquaredFactor
Sep 16, 2017