Slope-Intercept Form

Slope-Intercept Form exercise

• Describe the effects of changing $m$ or $b$.

  Hints

  Draw the lines of

  • $y=2x+2$ and
  • $y=4x+2$

  into the same coordinate system.

  What can you see? If the slope changes, what happens?

  The slope is:

  $m=\frac{\Delta y}{\Delta x}$.

  Solution

  The slope-intercept form is:

  $y=mx+b$, where

  • $m$ is the slope
  • $b$ is the y-intercept.

  $m$:

  • if $m$ is positive (negative), the line increases (decreases).
  • if the absolute value of $m$ increases (decreases), the line is steeper (less steep).
  • if $m=0$, the line is parallel to the x-axis.

  $b$:

  • this is the point where the line crosses the y-axis.
  • if $b$ increases (decreases), the line lies higher (lower) in the coordinate system.

• Decide whether or not the Mars rover has enough battery power left to drive eight miles.

  Hints

  The slope-intercept form is:

  $y=mx+b$, where

  • $m$ is the slope
  • $b$ is the y-intercept.

  For each mile the rover travels the battery power decreases by 2 units.

  Which means we'll have a negative slope.

  The x-intercept is at $y=0$.

  Solution

  The rover starts with 20 units of battery power left. This is the y-intercept. Because the rover needs two units of battery power each mile, the slope is $m=-2$. Each mile the battery power decreases by 2 units.

  This gives us the equation: $y=-2x+20$.

  We can determine the x-intercept of this equation:

  $\begin{array}{rcr} 0&=&-2x+20\\ \color{#669900}{+2x} && \color{#669900}{+2x}\\ 2x&=&20. \end{array}$.

  Now we divide by $2$ and get the x-intercept at: $x=10$.

  Therefore, the Mars rover has enough battery power left to drive eight miles ($8<10$).

• Determine the lines and corresponding equations.

  Hints

  The slope-intercept form is:

  $y=mx+b$

  • $m$ is the slope
  • $b$ is the y-intercept.

  First, have a look at the y-intercept.

  For the slope, calculate the change in $y$ divided by the change in $x$.

  $m=\frac{\Delta y}{\Delta x}$.

  Solution

  The y-intercept of the blue, red. and violet lines is at $2$. The other two lines have the y-intercept at $4$.

  Let's have a look at the slope:

  • The blue line: for each four steps to the right, we go one step down - this gives us the slope $m=-\frac14$. This is the equation: $y=-\frac14x+2$.
  • The yellow line is parallel to the blue line, so the lines have the same slope $m=-\frac14$. This is the equation: $y=-\frac14x+4$.
  • The red line: for each two steps to the right, we go one step up - this gives us the slope $m=\frac12$. This is the equation: $y=\frac12x+2$.
  • The green line is parallel to the red line. So the lines have the same slope $m=\frac12$. This is the equation: $y=\frac12x+4$.
  • The violet line: for each step to the right we go one step up - this gives us the slope $m=\frac11=1$. This is the equation: $y=x+2$.

• Determine the equations which the describe the Mars rover's expedition.

  Hints

  To write the slope-intercept form you need:

  • the slope $m$ and
  • the y-intercept $b$.

  The slope is the change in $y$ divided by the change in $x$:

  $m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$.

  Plug the x- and y-coordinates of the given points into the definition of $m$.

  Pay attention to the order of subtraction.

  For the intercept you can

  • use the point-slope form
  • substitute one point into the slope-intercept form

  Solution

  If you'd like to write the slope-intercept form of an equation with the two points ($P(4,0)$ and $Q(9,-7)$):

  1. Calculate the slope by $m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$
  2. Determine the y-intercept with the slope.

  The slope

  $m=\frac{\Delta y}{\Delta x}=\frac{-7-0}{9-4}=-\frac75=-1.4$.

  Now we can write down the equation as follows

  $y=-1.4x+b$.

  The y-intercept

  We substitute $x$ and $y$ with the given coordinates of one of the two points. It doesn't matter which point we choose. With the point $P(4,0)$ - $x=4$ and $y=0$, we get the equation

  $0=-1.4\times 4+b$.

  This gives us $b=5.6$.

  This leads to: $y=-1.4x+5.6$.

• Decide which equations are in slope-intercept form.

  Hints

  The slope indicates the steepness of the line.

  The y-intercept is the point where the line crosses the y-axis.

  The point-slope-form is:

  $y-y_1=m(x-x_1)$.

  Solution

  The slope-intercept form is:

  $y=mx+b$, where

  • $m$ is the slope. The greater the absolute value of the slope, the steeper the line.
  • $b$ is the y-intercept, the point where the line crosses the y-axis.

  • $y=-2x+20$ and $y=2x+1$ are in slope-intercept form.
  • $y-10=-2(x-5)$ is the point-slope form of $y=-2x+20$.
  • $m=\frac{\Delta y}{\Delta x}$ is the slope.

• Determine the correct equation for the given points.

  Hints

  $\Delta y=y_2-y_1$ is the change in $y$, and $\Delta x=x_2-x_1$ is the change in $x$.

  If the y-intercept is know, substitute the coordinates of any given point into the slope-intercept form to calculate the slope.

  You can also check each the equation, do both points satisfy the equation?

  Solution

  • First, we calculate the slope: $m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$
  • Then, we determine the y-intercept.

  $\mathbf{P(3,2)}$ and $\mathbf{Q(4,4)}$

  1. $m=\frac{\Delta y}{\Delta x}=\frac{4-2}{4-3}=\frac21=2$.
  2. Next we substitute the coordinates of, for example $Q$, in the equation $y=2x+b$. Rewrite the equation as $4=2\times 4+b$. Subtracting $8$ we get $b=-4$.

  This gives us the following equation $y=2x-4$.

  $\mathbf{P(1,-2)}$ and $\mathbf{Q(0,1)}$

  1. $m=\frac{\Delta y}{\Delta x}=\frac{1-(-2)}{0-1}=-\frac31=-3$.
  2. Next we substitute the coordinates of, for example $P$, in the equation $y=-3x+b$. Rewrite the equation as $-2=-3\times 1+b$. Adding $3$ we get $b=1$.

  This gives us the following equation $y=-3x+1$.

  $\mathbf{P(4,6)}$ and $\mathbf{Q(2,5)}$

  1. $m=\frac{\Delta y}{\Delta x}=\frac{5-6}{2-4}=\frac12=0.5$.
  2. Next we substitute the coordinates of, for example $P$, in the equation $y=0.5x+b$. Rewrite the equation as $6=0.5\times 4+b$. Subtracting $2$ we get $b=4$.

  This gives us the following equation $y=0.5x+4$.

  $\mathbf{P(-6,0)}$ and $\mathbf{Q(3,-3)}$

  1. $m=\frac{\Delta y}{\Delta x}=\frac{-3-0}{3-(-6)}=-\frac39=-\frac13$.
  2. Next we substitute the coordinates of, for example $Q$, in the equation $y=-\frac13x+b$. Rewrite the equation as $-3=-\frac13\times 3+b$. Adding $1$ we get $b=-2$.

  This gives us the following equation $y=-\frac13x-2$.