Write the following sentence by parts.

A number squared:  \(\displaystyle x^2\)

The difference of six and a number squared:  \(\displaystyle 6-x^2\)

Is four:  \(\displaystyle =4\)

Combine the parts to write an equation.

The answer is:  \(\displaystyle 6-x^2=4\)

This question is testing the knowledge and skills of calculating a function value. Similar to domain and range, calculating the function value requires the application of input values into the function to find the output value. In other words, evaluating a function at a particular value results in the function value; which is another way of saying function value is the output. 

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the input value.

Since the function is in the form \(\displaystyle f(x)\) and the question asks to calculate \(\displaystyle f(6)\) we know that the input value is six.

\(\displaystyle f(6)\rightarrow x=6\)

Step 2: Given the function, input the desired \(\displaystyle x\) value. 

In step one the input value of six was found. Now substitute 6 in for every\(\displaystyle x\) in the function.

\(\displaystyle f(x)=x^2-2x\)

\(\displaystyle f(6)=6^2-2(6)\)

\(\displaystyle f(6)=36-12\)

\(\displaystyle f(6)=24\)

Step 3: Verify solution by graphing the function.

To find the function value on a graph go to \(\displaystyle x=6\) and find the y value that lies on the graph at \(\displaystyle x=6\). Looking at the graph above when \(\displaystyle x=6\), \(\displaystyle y=24\).

Recall that \(\displaystyle f(x)=y \rightarrow f(6)=24\), thus verifying the solution found in step 2.

To solve this problem, first recall the equation of a line.

\(\displaystyle y=mx+b\)

where

\(\displaystyle \\m=\text{slope} \\y=\text{y-intercept}\)

Remember that slope is the rate of change that occurs in a function and that the \(\displaystyle y\)-intercept is the \(\displaystyle y\) value corresponding to an \(\displaystyle x\) value equalling zero.

Now, identify what is known.

"Billy buys a tomato plant that is 4 inches tall. With regular watering the plant grows 3 inches a year."

Since the height of Billy's plant is 4 inches tall when he gets it, at time being zero. It is also stated that the plant grows 3 inches per year, this is the rate of change of the plant's height. Writing the known information in mathematical terms results as follows.

\(\displaystyle h(t)=3t+4\)

In this case our 

\(\displaystyle \\y=h(t)=\text{plant's height as a function of time} \\t=years \\m=3 \\b=4\)

Therefore, the \(\displaystyle y\)-intercept represents the starting height of the plant which is 4 inches.

To solve this problem, first recall the equation of a line.

\(\displaystyle y=mx+b\)

where

\(\displaystyle \\m=\text{slope} \\y=\text{y-intercept}\)

Remember that slope is the rate of change that occurs in a function and that the \(\displaystyle y\)-intercept is the \(\displaystyle y\) value corresponding to an \(\displaystyle x\) value equalling zero.

Now, identify what is known.

"Billy buys a tomato plant that is 4 inches tall. With regular watering the plant grows 3 inches a year."

Since the height of Billy's plant is 4 inches tall when he gets it, at time being zero. It is also stated that the plant grows 3 inches per year, this is the rate of change of the plant's height. Writing the known information in mathematical terms results as follows.

\(\displaystyle h(t)=3t+4\)

In this case our 

\(\displaystyle \\y=h(t)=\text{plant's height as a function of time} \\t=years \\m=3 \\b=4\)

Therefore, the slope represents the rate of change in the height of the plant which is 3 inches per year.

First calculate the slope of the line using the slope formula.

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

Substituting in the known information.

\(\displaystyle \\(x_1, y_1)=(1,2) \\(x_2,y_2)=(4,5)\)

\(\displaystyle m=\frac{y_2-y_1}{x_2-x_1}=\frac{5-2}{4-1}=\frac{3}{3}=1\)

Now use point slope form to find the equation of the line passing through these points.

\(\displaystyle \\y-y_1=m(x-x_1) \\y-2=1(x-1) \\y-2=x-1 \\y=x+1\)

Now identify which point lies on the line.

\(\displaystyle \\(9,10)\Rightarrow10=9+1 \\(3,2)\Rightarrow 2\neq3+1 \\(0,0)\Rightarrow0\neq0+1 \\(2,2)\Rightarrow 2\neq2+1 \\(5,3)\Rightarrow 3\neq5+1\)

Therefore, the point that lies on the line is (9,10)

To solve this problem first manipulate the given equation to solve for \(\displaystyle c\).

\(\displaystyle \frac{c}{d}=6\)

Multiply by \(\displaystyle d\) on both sides. The \(\displaystyle d\)'s on the left hand side of the equation will cancel out.

\(\displaystyle d\cdot\frac{c}{d}=6\cdot d\)

\(\displaystyle c=6d\)

Now to calculate \(\displaystyle \frac{3d}{c}\), substitute the value for \(\displaystyle c\) that was just found, into the denominator and simplify by canceling out common factors.

\(\displaystyle \frac{3d}{c}=\frac{3d}{6d}=\frac{3}{6}=\frac{1\cdot 3}{2\cdot 3}=\frac{1}{2}\)

\(\displaystyle y=x^2-5x+6\)

First we group terms

\(\displaystyle y=x^2-5x+6=(x^2-5x)+6\)

Now we want to have a perfect square, so we add \(\displaystyle \left(\frac{b}{2}\right)^2\), and we subtract \(\displaystyle \left(\frac{b}{2}\right)^2\), so now it looks like

\(\displaystyle \left(x^2-5x+\left(\frac{-5}{2}\right)^2\right)-\left(\frac{-5}{2}\right)^2+6\)

\(\displaystyle (x^2-5x+6.25)-6.25+6\)

Simplify to get

\(\displaystyle y=(x-2.5)^2-0.25\)

\(\displaystyle f(x) = x^2 + 32x -12\)

\(\displaystyle f(4) = 4^2 + 32(4) -12\)

\(\displaystyle = 16 + 128 -12\)

\(\displaystyle = 132\)

We can use point slope form to determine the equation of a line that fits the data. 

Point slope form is \(\displaystyle y-y_0=m(x-x_0)\), where \(\displaystyle (x_0, y_0)\), and \(\displaystyle m\) is the slope, where \(\displaystyle m=\frac{y_2-y_1}{x_2-x_1}\).

Let \(\displaystyle y_2=6\), \(\displaystyle y_1=3\), \(\displaystyle x_2=3\), and \(\displaystyle x_1=2\).

\(\displaystyle m=\frac{6-3}{3-2}=3\)

If we do this for every other point, we will see that they have the same slope of \(\displaystyle 3\).

\(\displaystyle y-y_0=3(x-x_0)\)

Now let \(\displaystyle x_0=0\), and \(\displaystyle y_0=-3\).

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

\(\displaystyle y+3=3x\)

\(\displaystyle y=3x-3\)

If we use the equation that solved for previously, \(\displaystyle y=5x+2\), we just plug in \(\displaystyle 15\) for \(\displaystyle x\).

\(\displaystyle y=5(15)+2=75+2=77\)