[br][br]A function is said to be a linear function if there is only one variable with the exponent one:[br][br][br] [math]\LARGE f(x) = kx + c,[/math][br][br] [br]where [i]k[/i] is the gradient (the slope of the function) and [i]c[/i] is a vertical intercept. The graph of a linear function is a straight line which is NOT parallel to[i] x[/i]-axis ([math]\large k\ne0[/math]).[br][br]As it can be seen from the applet, the sign of the gradient defines if the linear graph is ascending or descending:[br][br] k > 0: the linear graph is ascending[br][br] k < 0: the linear graph is descending. [br][br]

If one point [math]\Large (x_0,\; y_0)[/math] and the gradient [i]k[/i] are known, the linear function is uniquely defined. The linear function can be defined by the formula[br][br] [math]\huge\textcolor{blue}{y-y_0=k(x-x_0).}[/math][br] [br][br][u][color=#0000ff]Example 4[/color].[/u] Define the linear function which is passing through the point (-2, 1) and has the gradient -3.[br][br]The known point ([i]x[/i][sub]0[/sub], [i]y[/i][sub]0[/sub]) is (-2, 1) and the gradient [i]k[/i] = -3. By substituting this to the formula, we get [br][br] [math]\Large\begin{eqnarray} &y-y_0&=k(x-x_0)\\[br]&y-1&=-3[(x-(-2)]\\[br]\Leftrightarrow& y-1&=-3(x+2)\\[br]\Leftrightarrow& y-1&=-3x-6\\[br]\Leftrightarrow& y&=-3x-5\end{eqnarray}[/math][br][br] [br]The straight line between two points is unique and the linear function can be defined with two points. The gradient is the slope of the line: the greater gradient (without the sign) the steeper slope.[br][br]

When defining the line with two points (x[sub]1[/sub], y[sub]1[/sub]) and (x[sub]2[/sub], y[sub]2[/sub]), Δy and Δx are obtained with the points:[br] [br] [math]\LARGE \textcolor{blue}{k=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}.}[/math][br][br] [br][color=#0000ff][br]Example 2.[/color] Define the line going through points (-2, 1) and (3, 5). [br] [br][color=#0000ff]Solution 2[/color]. The known points are [math]\large (x_1,\, y_1) = (-2, \,1)[/math] and [math]\large (x_2,\, y_2) = (3, \,5)[/math]. Thus, the slope si[br][br] [math]\Large k=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}=\frac{5-1}{3-(-2)}=\frac{4}{5}.[/math][br] [br]The equation of the line is[br] [br] [math]\Large \begin{eqnarray}[br]&y-5&=\frac{4}{5}(x-3)\\[br] \Leftrightarrow & y-5&=\frac{4}{5}-\frac{12}{5}\\[br] \Leftrightarrow &y&=\frac{4}{5}x-\frac{12}{5}+5\\[br] \Leftrightarrow &y&=\frac{4}{5}x+\frac{13}{5}\end{eqnarray}[/math] [br] [br] [br]The equation of the line can be also given is the standard form [math]\Large 4x - 5y + 13 = 0. [/math] As a function, it would be written as[br][br] [math]\Large f(x)=\frac{4}{5}x+\frac{13}{5}.[/math][br][br][br][color=#0000ff]Example 3.[/color] The pressure at the beginning was 6.0 MPa and one minute later 0 MPa. If the pressure is linearly dependent on time, what was the pressure at [i]t[/i] = 32 [i]s[/i].[br][br]Known points are [math]\Large (t_0,\,p_0)=(0,\,6)[/math] ja [math]\Large (t_1,\,p_1)=(60,\,0).[/math] [br][br]That is, [br][br] [math]\Large \begin{eqnarray}[br]k&=&\frac{\Delta p}{\Delta t}=\frac{6-0}{0-60}=-\frac{1}{10}\\[br]\vspace{5mm}\\[br]p-p_0&=&k(t-t_0)\\[br]p-6&=&-\frac{1}{10}(t-0)\\[br]p&=&-\frac{1}{10}t+6.[br]\end{eqnarray}[/math][br] [br][br]Using the equations as a function, we can substitute given time [i]t[/i] = 32 [i]s[/i] and the pressure seems to be [br] [br] [math]\Large p(32)=-\frac{1}{10}\cdot 32+6=2.8\;\text{ (MPa)}[/math][br] [br]

[br]1. If the point is given, use it as a starting point. If a function is known, use the intercept as a starting point. [br][br]2. Go to the right from [br]the starting point by the denominator of the gradient and up or down by the numerator of the gradient. [br][br]3. If the gradient is negative, go down and vice versa. This is your second point.Draw the linear graph through these points. [br]