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All courses Algebra I · F-IF.7.a 67 of 76
Graph linear and quadratic functions and identify intercepts, maxima, and minima.

Use slope-intercept form to list graphing features of a line

Problem
Extract graphing features from \(y=2x+3\).
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Hint

Read m and b separately, then express m as an integer rise over a positive run.

In y=mx+b, intercept=(0,b) and m=rise/run. Choose a positive run and carry the slope sign in the rise.

Solution walkthrough

01

Match slope-intercept form

\[\begin{aligned} y=mx+b \\ y=2x+3~->~m=2=2/1;~b=3 \end{aligned}\]

In slope-intercept form, m is the coefficient of x and b is the output when x=0.

02

Write the first graph point

\[b=3~->~y-\text{intercept}~(0,3)\]

The line crosses the y-axis where x=0. Substitution gives y=2(0)+3=3, so the point is (0,3).

03

Use the slope as a move

\[\begin{aligned} m=\text{rise}/\text{run}=2/1 \\ \text{from}~(0,3):~\text{right}~1,~\text{up}~2~->~(1,5) \end{aligned}\]

A positive slope of 2 means that each one-unit increase in x produces a two-unit increase in y.

04

Verify the second point

\[\begin{aligned} x=1:~y=2(1)+3=5 \\ 5=2*1+3 \end{aligned}\]

The computed output agrees with the moved point (1,5), so the intercept and slope features are mutually consistent.

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Another way

  1. Evaluate the equation at x=0 and x=1 to get (0,3) and (1,5); their slope is (5-3)/(1-0)=2.

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Common mistake

Do not treat 3 as the slope. In y=mx+b, the x-coefficient 2 is the slope and the constant 3 is the y-intercept output.

Answer graph labeling every exact line or quadratic feature required by the variant.
slope field=m=2=2/1; y-intercept field=(0,3); rise/run move=right1,up2; second point=(1,5); substitution check=5=2·1+3

Solution walkthrough video