6.3: Solving Graphically and Algebraically | Algebra 2 Honors

How it works

An intersection solves the equality between two function outputs. A graph helps locate the number and approximate positions of intersections, while algebra can provide exact values. Check each result in the original equation and apply the context domain before accepting it.

f(x)=g(x)\quad\Longleftrightarrow\quad\text{common points of the graphs}

Worked examples

Example 1

Solve $\displaystyle x^2=x+6$ .

Move terms: $\displaystyle x^2-x-6=0=(x-3)(x+2)$ . Thus $\displaystyle x=-2$ or $\displaystyle x=3$ .

Example 2

Solve $\displaystyle 2x+5=17$ and interpret it as an intersection.

$\displaystyle 2x=12$ , so $\displaystyle x=6$ . The lines $\displaystyle y=2x+5$ and $\displaystyle y=17$ meet at $\displaystyle (6,17)$ .

Problems

Problem 6.3.1

Solve $\displaystyle x^2=9-x^2$ .

Show worked solution

$\displaystyle 2x^2=9$ , so $\displaystyle x=\pm3/\sqrt{2}=\pm3\sqrt{2}/2$ .

Problem 6.3.2

Solve $\displaystyle \sqrt{x+2}=x$ and state why both squared-equation roots are not valid.

Show worked solution

Squaring gives $\displaystyle x=-1$ or $\displaystyle x=2$ . The original right side must be nonnegative; only $\displaystyle x=2$ checks.

Problem 6.3.3

Use the graphs of $\displaystyle y=x^2$ and $\displaystyle y=x+2$ to find their intersections, then verify algebraically.

The graphs y = x squared and y = x + 2 intersect twice. — Each common point solves both equations.

Show worked solution

From $\displaystyle x^2=x+2$ , factor $\displaystyle (x-2)(x+1)=0$ . The intersections are $\displaystyle (-1,1)$ and $\displaystyle (2,4)$ .

The graphs y = x squared and y = x + 2 intersect twice. The intersections are (−1, 1) and (2, 4). — The intersections are (−1, 1) and (2, 4).

Lesson Problems PDF Solutions PDF Unit MCQ

How it works

Worked examples

Example 1

Example 2

Problems

Problem 6.3.1

Problem 6.3.2

Problem 6.3.3

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