Gradient-intercept form
Isolate the vertical variable, then read the signed coefficient and constant. The intercept value is the vertical coordinate of the crossing
- Gradient The coefficient includes its sign, wherever the term is written \(\displaystyle m\)
- Vertical-axis crossing The horizontal coordinate is zero at this crossing \(\displaystyle (0,c)\)
- Through the origin A missing constant is zero \(\displaystyle y=mx\)
- Horizontal line A missing horizontal-variable term has coefficient zero \(\displaystyle y=c\)
- Vertical line Its gradient is undefined. A vertical line away from the vertical axis has no vertical-axis crossing. The vertical axis itself has no unique crossing with that axis \(\displaystyle x=k\)
Worked example
Find the gradient and vertical-axis intercept
- \(\displaystyle 4y=20-7x\)Subtract seven times the horizontal variable from both sides
- \(\displaystyle y=\frac{20}{4}-\frac{7}{4}x\)Divide every term by four
- \(\displaystyle y=5-\frac{7}{4}x\)Simplify the constant
- \(\displaystyle m=-\frac{7}{4}\)Read the signed coefficient
- \(\displaystyle c=5\)Read the constant for the vertical-axis intercept
Common mistakes
Difficulty levels
Questions about this topic
Does the term order matter?+
No. Identify the signed coefficient and constant by their roles
Is the intercept a number or a point?+
The intercept value is the vertical coordinate. To name the crossing as a point, put zero first and that value second
What changes when the gradient changes?+
With the intercept fixed, the line turns through its vertical-axis crossing. Its direction and steepness can change
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 11, VCE Units 1 & 2, General Mathematics, chapter 5: Linear functions and models