x²   + 2x - 1 = 7

x²   + 2x - 8 = 0

(x + 4) (x - 2) = 0

x = -4, x = 2

The vertex form of a parabola is given by the equation:

\(\displaystyle f(x)=a(x-h)^2 +k\), where the point \(\displaystyle \dpi{100} (h,k)\) is the vertex, and \(\displaystyle \dpi{100} a\) is a constant.

We are told that the vertex must occur at \(\displaystyle \dpi{100} (3,4)\), so let's plug this information into the vertex form of the equation. \(\displaystyle \dpi{100} h\) will be 3, and \(\displaystyle \dpi{100} k\) will be 4.

\(\displaystyle f(x)=a(x-3)^2 +4\)

Let's now expand \(\displaystyle (x-3)^2\) by using the FOIL method, which requires us to multiply the first, inner, outer, and last terms together before adding them all together.

\(\displaystyle (x-3)^2 = (x-3)(x-3)=x^2-3x-3x+9=x^2-6x+9\)

We can replace \(\displaystyle (x-3)^2\) with \(\displaystyle x^2-6x+9\).

\(\displaystyle f(x)=a(x-3)^2+4=a(x^2-6x+9)+4\)

Next, distribute the \(\displaystyle \dpi{100} a\).

\(\displaystyle a(x^2-6x+9)+4 = ax^2 -6ax+9a+4\)

Notice that in all of our answer choices, the first term is \(\displaystyle -2x^2\). If we let \(\displaystyle \dpi{100} a=-2\), then we would have \(\displaystyle -2x^2\) in our equation. Let's see what happens when we substitute \(\displaystyle \dpi{100} -2\) for \(\displaystyle \dpi{100} a\).

\(\displaystyle f(x)=ax^2-6ax+9a+4=(-2)x^2-6(-2)x+9(-2)+4\)

\(\displaystyle =-2x^2+12x-18+4\)

First, we set the quadratic function equal to \(\displaystyle 0\):

\(\displaystyle x^{2}-12x+36=9\)

\(\displaystyle x^{2}-12x+27=0\)

Reduce the function to its two component factors:

\(\displaystyle (x-3)(x-9)=0\)

Therefore, since either \(\displaystyle x-3=0\) or \(\displaystyle x-9=0\),

\(\displaystyle x=3, x=9\)

First, set the quadratic function equal to \(\displaystyle 0\):

\(\displaystyle x^{2}+5x-3=-9\)

\(\displaystyle x^{2}+5x+6=0\)

Then, reduce the function to its two factors:

\(\displaystyle (x+3)(x+2)=0\)

Since one of the factors on the left hand side of the equation must equal \(\displaystyle 0\) in order for the above equation to be true,

\(\displaystyle x+3=0\) or \(\displaystyle x+2=0\)

Solving for both, we get \(\displaystyle x=3, x=2\).

First, set the quadratic function equal to \(\displaystyle 0\):

\(\displaystyle x^{2}+x-50=-8\)

\(\displaystyle x^{2}+x-42=0\)

Then, separate the function into its two component factors:

\(\displaystyle (x-6)(x+7)=0\)

It follows from this equation that either \(\displaystyle x-6=0\) or \(\displaystyle x+7=0\), 

\(\displaystyle x=6, x=-7\)

The area of a rectangle is the product of its length by its width, which we know to be equal to \(\displaystyle 125\:ft^2\) in our problem. We also know that the length is equal to \(\displaystyle 2+3w\), where \(\displaystyle w\) represents the width of the land. Therefore, we can write the following equation:

\(\displaystyle w(2+3w)=125\)

Distributing the \(\displaystyle w\) outside the parentheses, we get:

\(\displaystyle 3w^{2}+2w=125\)

Subtracting \(\displaystyle 125\) from each side of the equation, we get:

\(\displaystyle 3w^{2}+2w-125=0\)

We get a quadratic equation, and since there is no factor of \(\displaystyle 3\) and \(\displaystyle -125\) that adds up to \(\displaystyle 2\), we use the quadratic formula to solve this equation.

\(\displaystyle a=3\)     \(\displaystyle b=2\)      \(\displaystyle c=-125\)

\(\displaystyle w=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)

We can first calculate the discriminant (i.e. the part under the square root)

\(\displaystyle b^{2}-4ac= 2^{2}-4\times3 \times(-125)\)

\(\displaystyle 4-4\times3\times(-125)\)

\(\displaystyle 4-12\times(-125)\)

\(\displaystyle 4-(-1500)=1504\)

We replace that value in the quadratic formula, solving both the positive version of the formula (on the left) and the negative version of the formula (on the right):

\(\displaystyle w=\frac{-2+\sqrt{1504}}{6}\)                                                              \(\displaystyle w=\frac{-2+\sqrt{1504}}{6}\)

Breaking down the square root:

\(\displaystyle w=\frac{-2+\sqrt{2\times2\times2\times2\times2\times 47}}{6}\)                          \(\displaystyle w=\frac{-2-\sqrt{2\times2\times2\times2\times2\times 47}}{6}\)

We can pull two of the twos out of the square root and place a \(\displaystyle 4\) outside of it:

\(\displaystyle w=\frac{-2+4\sqrt{2\times 47}}{6}\)                  \(\displaystyle w=\frac{-2-4\sqrt{2\times 47}}{6}\)

We can then multiply the \(\displaystyle 2\) and the \(\displaystyle 47\):

\(\displaystyle w=\frac{-2+4\sqrt{94}}{6}\)                                   \(\displaystyle w=\frac{-2-4\sqrt{94}}{6}\)

At this point, we can reduce the equations, since each of the component parts of their right sides has a factor of \(\displaystyle 2\):

\(\displaystyle w=\frac{-1+2\sqrt{94}}{3}\)                                   \(\displaystyle w=\frac{-1+2\sqrt{94}}{3}\)

Since width is a positive value, the answer is:

\(\displaystyle w=\frac{-1+2\sqrt{94}}{3}\)

\(\displaystyle w=\frac{-1+2\times9.695}{3}\)

\(\displaystyle w=\frac{-1+19.39}{3}\)

\(\displaystyle w=\frac{18.39}{3}=6.13\)

 The width of the piece of land is approximately \(\displaystyle 6.13\:ft^2\).

The focus is solved by using the following formula \(\displaystyle (x,y+p)\), where \(\displaystyle x,y\) are the coordinates of the vertex, and \(\displaystyle p\) is the distance from the vertex. To solve for \(\displaystyle p\), we use

\(\displaystyle a=\frac{1}{4p}\), where \(\displaystyle a\) is the coefficient in front of the \(\displaystyle x^2\) term and \(\displaystyle p\) is the focus.

Since \(\displaystyle a=2\), we can substitute in to get,

\(\displaystyle 2=\frac{1}{4p}\)

\(\displaystyle 2p=\frac{1}{4}\)

\(\displaystyle p=\frac{1}{4}\cdot\frac{1}{2}=\frac{1}{8}\)

Now we need to find the vertex of the equation.

Since our equation is in vertex form, we can deduce that the vertex is at \(\displaystyle (3,5)\)

So the focus is at \(\displaystyle \left( 3,5+\frac{1}{8}\right)=\left( 3,\frac{41}{8}\right)\)

In order to find all the solutions, we need to set the equations equal to each other.

\(\displaystyle x^3+1=x+1\)

Now subtract \(\displaystyle x\) and \(\displaystyle 1\) from each side.

\(\displaystyle x^3-x+1-1=0\)

\(\displaystyle x^3-x=0\)

Factor the left hand side to get

\(\displaystyle x(x^2-1)=0\)

Factor the quadratic function inside the parenthesis to get

\(\displaystyle x(x-1)(x+1)=0\)

The solutions to this equation are

\(\displaystyle x=0\)

\(\displaystyle x-1=0\rightarrow x=1\)

\(\displaystyle x+1=0\rightarrow x=-1\)

To figure out how many x-intercepts there are, we need to set out equation equal to zero.

\(\displaystyle f(x)=x^3-x\)

\(\displaystyle 0=x^3-x\)

Factor an x out

\(\displaystyle 0=x(x^2-1)\)

Factor inside of the parenthesis.  

\(\displaystyle 0=x(x-1)(x+1)\)

We can see there are 3 x-intercepts at \(\displaystyle x=0, x=1, x=-1\).

If the equation 

\(\displaystyle C=1.2+.5h\) 

were graphed, the result would be a line with a slope of 0.5 and a y-intercept of 1.2. 0.5 represents the amount Tonya pays per hour to work at a computer, because we know that the rate is $0.50, and C represents the total amount she pays. Because we must add 1.2 (or $1.20) to the total amount of hours multiplied by the hourly rate in order to get the total amount Tonya paid, we can assume there is another charge for renting a computer, in addition to the hourly rate, and that that charge is $1.20. Therefore, the y-intercept of the graph of the equation represents the initial cost of using a computer.