Choose the correct branch of a modulus graph

The Teacher completes the line $y=x-1$ on the Student sketch. The modulus curve has zeros $x=\pm\sqrt3$ and height 3 at 0. Outside $[-\sqrt3,\sqrt3]$ it follows $y=x^2-3$; inside it follows $y=-(x^2-3)=3-x^2$. The source labels the right-hand intersection $A$ on the original upward branch and $B$ on the reflected central section. The line crosses the $x$-axis at 1.

The source’s “Equivalence” note connects algebraic and graphical representations. Independently $|x^2-3|\ge0$ requires $x-1\ge0$ at any intersection, so $x\ge1$. This rules out intersections on the negative side before solving.

The source’s complete algebra is

$x-1=x^2-3,\qquad x^2-x-2=0,\qquad (x+1)(x-2)=0.$

This gives $x=-1$ or 2. The source rejects −1 using the diagram. Independently, the branch requires $|x|\ge\sqrt3$ and every intersection requires $x\ge1$, so −1 is invalid. At $x=2$, $y=x-1=1$, giving $A=(2,1)$.

Both copies give

$x-1=-(x^2-3),\qquad x^2+x-4=0,$

then the full quadratic-formula chain

$\begin{aligned}&x\\ & =\frac{-1\pm\sqrt{1-4(1)(-4)}}2\\ & =\frac{-1\pm\sqrt{17}}2.\end{aligned}$

The diagram rejects $(-1-\sqrt{17})/2$. Independently it is negative and outside the active central interval; the valid root $(-1+\sqrt{17})/2$ lies between 1 and $\sqrt3$. Substituting into $y=x-1$ gives $(-3+\sqrt{17})/2$. Thus

$B=\left(\frac{-1+\sqrt{17}}2,\frac{-3+\sqrt{17}}2\right).$

The source warns to use the correct pair of Cartesian equations when finding intersections involving a modulus. The two disjoint active regions above account for every possible intersection.

The source’s six-step procedure begins on 26:

1. Enter $y=x-1$ in Y₁ and $y=|x^2-3|$ in Y₂. For modulus, use MATH → right-arrow → 1:abs(, or ALPHA → WINDOW → 1:abs(. Group the whole quadratic inside abs.
2. Select ZOOM → 6:ZStandard, giving $-10\le x\le10$ and $-10\le y\le10$.
3. Select ZOOM → 2:Zoom In, move the cursor near the two intersections and press ENTER to enlarge that portion.
4. Press 2nd → TRACE → 5:intersect. The menu also shows value, zero, minimum, maximum, dy/dx and integral operations; intersect is the operation needed here.

The reconstructed screens retain the two entries, standard/enlarged views and selected menu. The enlarged reconstruction uses explicitly chosen bounds, not claimed original settings.

Step 5 continues across 26–27. With First curve? showing Y₁, move near $A$ and press ENTER. With Second curve? showing Y₂, press ENTER. At Guess?, press ENTER again. The result appears at the bottom as Intersection: X=2, Y=1.

The source’s intermediate cursor is approximately $X=2.0833333$, with Y₁≈1.0833333 and Y₂≈1.3402778. These are cursor/guess readings before the solver result, not an additional intersection. The final A result agrees with the algebra.

Step 6 repeats the same sequence nearer $B$: select Y₁ at First curve? and press ENTER; select Y₂ at Second curve? and press ENTER; at Guess?, press ENTER. The source’s intermediate $X\approx1.5151515$ has Y₁≈0.5151515 and Y₂≈0.7043159. The solver’s final screen gives $X\approx1.5615528$, $Y\approx0.5615528$.

The closing source remark requires algebraic/analytical working whenever exact values are requested; a graphing-calculator result is non-exact. The exact radical result in the preceding part agrees with the rounded screen, and should be retained as the final mathematical answer.