Graphing equations and finding intersections

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Turn an equation into a picture, then turn the picture back into mathematics.

By the end of this section

Find roots, intersections, and vertices; choose useful viewing bounds; preserve original restrictions; distinguish a touching point from a crossing; and recognize when a graph does not establish a complete answer.

2.1 Translate the question into the correct graph

For an equation $L(x) = R(x),$ graph $y = L(x)$ and $y = R(x).$ At every intersection, both expressions have the same output. The x-coordinate is a candidate solution of the original one-variable equation.

Alternatively, graph $y = L(x) - R(x)$ and look for its x-intercepts. At a root, the difference is zero. The first approach preserves the two sides visibly; the second can make a useful graph window easier to choose.

Question	Graph setup	Read
Solve $f(x) = 0$	$y = f(x)$	Root’s x-coordinate.
Solve $f(x) = k$	$y = f(x)$ and $y = k$	Intersection’s x-coordinate.
Solve a system in x, y	Both original equations	Ordered pair $(x, y).$
Find the maximum of f	$y = f(x),$ on its domain	Highest output, not the input.
Find when a maximum occurs	Same graph	Input at the highest output.

Select the mathematical feature, not a nearby pixel

In Desmos, select a curve or its expression to reveal available points of interest. Select the relevant point to display its coordinate label. Intercepts and intersections can appear as these points.^[5] A point you happened to click on a curve is not automatically a root, vertex, or intersection.

If no useful label appears, check the entry and window. Then use an equivalent graph or algebra rather than endlessly clicking. A graphical solver is a numerical aid; the coordinates still need interpretation and verification.

The target line on your scratch paper

Before entering an equation, write a short target such as “larger x,” “the y-coordinate,” “maximum value,” or “ $2x + y.$ ” After obtaining a point (a, b), read that line again. Do not let the display choose the requested quantity for you.

2.2 A useful window is part of the solution

A graph window shows a bounded region of the coordinate plane. It is not the function’s domain. An empty window does not establish that an equation has no solution.

Set lower and upper bounds for both axes in Graph Settings, or pan and zoom to investigate. The table’s Zoom Fit control is useful for plotted data.^[6] A reasonable window comes from the problem: times may run from 0 to 20, while prices may run from 0 to 2,000. The axes do not need identical numerical bounds.

Plan the window from the mathematics

Estimate a landmark. For $x 2 = 625,$ roots are near $\pm25;$ a window restricted to $-10 \leq x \leq 10$ misses both.

Include the required output. For $y = 2x + 400$ and $y = -3x + 900,$ the intersection has a large positive output. A tiny vertical range can hide the important point even when the horizontal range is correct.

Recenter, then refine. Find the relevant region with broad bounds, then tighten the bounds around nearby roots or a vertex. Do not interpret the width of a thick line as the size of a mathematical interval.

Original explanatory graph. Both solutions lie beyond a narrow horizontal window centered near zero.

A different graph can be better than a bigger window

If both sides have very large common terms, subtract them first. Solving $1000 + x 2 = 1009$ can be viewed as the much simpler $x 2 - 9 = 0.$ Equivalent algebra reduces graph setup and makes exact checking easier.

2.3 Respect domains, boundaries, and multiplicity

A candidate intersection must satisfy the original expression. Exclude zero denominators, require nonnegative radicands for even real roots, and apply context restrictions such as $t \geq 0$ or an integer count.

Desmos supports curly-brace restrictions appended to a graph, such as $y=x^2{x>=0}.$ Strict inequalities use dotted boundaries; non-strict inequalities use solid boundaries.^[12] Use a graph to display the restriction you already understand, not to invent a restriction from the picture.

Three situations that a quick picture can blur

Situation	Required reasoning
A hole	The cancelled factor’s zero remains excluded from the original expression. A tiny missing point may not be visible.
A tangent contact	Two graphs can meet without crossing. One touching point is still one solution of the system.
Nearly equal roots	Two distinct roots may look like one at a coarse scale. Exact algebra or a closer window distinguishes them.

A repeated root is one distinct solution

For $(x - 2) 2 = 0,$ the graph touches the axis at (2, 0). The root has multiplicity 2 as a polynomial factor, but there is only one distinct real value of x satisfying the equation. A method that looks only for sign changes would miss this root.

Counting solutions requires more than counting visible dots

Use the structure: a nonzero quadratic has at most two distinct real roots; a line tangent to a circle has one intersection; a forbidden denominator value never becomes a solution after cancellation. If the problem asks for an exact count, justify why other solutions are excluded.

2.4 A repeatable graph-solving workflow

1. Define the target. Input, output, ordered pair, count, or expression of the coordinates?

2. Preserve the equation. Enter both original sides or a valid difference. Record restrictions first.

3. Set a useful view. Include expected inputs and outputs; move beyond a default window when necessary.

4. Read the right feature. Select the relevant intercept, intersection, or extremum. Record the whole point on scratch paper.

5. Verify and report. Substitute into the original conditions; reject forbidden values; compute the requested final quantity.

An equation can be graphed without isolating y

For a system such as $2x + y = 11$ and $x - y = 1,$ enter both equations directly. For a circle, enter its complete equation. Rearranging is optional when the calculator supports the implicit relation; it is still valuable for understanding how many intersections to expect.

Be explicit about variable roles. If a parameter h is the unknown in $(5 - h) 2 + 2 = 11,$ you can temporarily graph $y = (5 - x) 2 + 2$ and $y = 11,$ treating the graph’s horizontal coordinate as h. Write that substitution down. Otherwise the display’s letter x can obscure what the answer represents.

When a graph should trigger algebra

Use algebra to settle exact tangency, identities, excluded values, very close roots, and parameter conditions. Use the graph to discover or check a candidate, but do not treat a near match as an exact proof. In some plots, detail may not be fully resolved; Desmos itself documents limitations in rendering implicit equations.^[24]

Two efficient recovery moves

Nothing appears: inspect syntax and definitions, then estimate the relevant scale. Try plotting the difference of the two sides.

A point appears but the answer is unclear: write (x, y) with the actual coordinates and reread the target. A missing interpretation step is not repaired by more zooming.

What to practice before timing yourself

Find both roots of a quadratic, a system’s intersection, a vertex, and a point outside your initial window. Then repeat with the question asking for the other coordinate or for a simple expression such as $x + y.$ You should be able to explain every final number without pointing vaguely at a curve.

15 worked examples

2.01. Use an intersection to solve one equation

Solve 3x + 7 = 22.

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Recognize. At a solution, the two sides have the same output. Graph each side against the same input.

Enter: $y=3*x+7$

Enter: $y=22$

Work. The line and the horizontal line intersect at (5, 22). The required unknown is the input, so read the x-coordinate, 5. A useful view includes $0 \leq x \leq 8$ and $0 \leq y \leq 30.$

Answer: $x = 5.$

Check. $3(5) + 7 = 22.$ Algebra gives the same result in one short step: $3x = 15.$

Avoid the trap. The intersection's output 22 was supplied by the equation. It is not the value of x. This simple problem teaches the graph method; on a timed test, direct algebra is usually shorter.

2.02. Read the requested coordinate of a system

The system 2x + y = 11 and x − y = 1 has solution (x, y). What is y?

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Recognize. Each point on a graph satisfies one equation; a common point satisfies both.

Enter: $2*x+y=11$

Enter: $x-y=1$

Work. Graph both equations in a view containing the first quadrant. The intersection is (4, 3). Because the question asks for y, report 3 rather than 4 or the entire ordered pair.

Answer: 3.

Check. $2(4) + 3 = 11$ and $4 - 3 = 1.$ Both equations must hold.

Avoid the trap. Select the actual intersection, not an intercept of one line. Checking only one equation cannot verify a solution to the system.

2.03. Find both roots before applying a qualifier

What is the larger solution of x² − 5x + 6 = 0?

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Recognize. Roots are the input values where the graph has output zero.

Enter: $y=x^2-5*x+6$

Work. The parabola has x-intercepts (2, 0) and (3, 0). The word “larger” selects 3. To distinguish the close intercepts, a view such as $0 \leq x \leq 5$ and $-1 \leq y \leq 7$ is more useful than a very wide window.

Answer: 3.

Check. Factoring gives $(x - 2)(x - 3) = 0,$ so the complete solution set is {2, 3}.

Avoid the trap. Do not stop at the first visible root. A qualifier such as larger, smaller, positive, or negative is part of the target, not an optional detail.

2.04. Separate the maximum value from its location

The function h(t) = −2(t − 4)² + 18 models height for 0 ≤ t ≤ 7. What is its maximum value?

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Recognize. The coefficient of the square is negative, so the vertex gives a maximum.

Enter: $y=-2*(x-4)^2+18$

Work. Using x for the model's input t, the vertex is (4, 18). It lies within the stated time interval. The maximum height is the output 18; the time when it occurs is 4.

Answer: 18.

Check. Since $(t - 4) 2 \geq 0,$ $-2(t - 4) 2 + 18 \leq 18,$ with equality at $t = 4.$ This establishes the maximum exactly without relying on a pixel.

Avoid the trap. The highest value visible in a poorly chosen window is not necessarily the function's maximum. Include the vertex and check the allowed interval.

2.05. Solve an output equation rather than evaluate the output as an input

Let f(x) = x² − 4x − 1. What positive value of x satisfies f(x) = 4?

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Recognize. The number 4 is an output target. It is not the input for f(4).

Enter: $f(x)=x^2-4*x-1$

Enter: $y=4$

Work. The intersections occur at (−1, 4) and (5, 4). The requested positive input is 5. A view from $x = -2$ to $x = 6$ shows both candidates.

Answer: 5.

Check. $f(5) = 25 - 20 - 1 = 4.$ In contrast, $f(4) = 16 - 16 - 1 = -1,$ which answers a different question.

Avoid the trap. Function notation is not an instruction to plug in every number you see. Decide whether the given number names an input or a required output before entering it.

2.06. Keep both branches of an absolute-value equation

Find all real solutions of |2x − 3| = 7.

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Recognize. An absolute value can equal 7 when its inside expression is either 7 or −7.

Enter: $y=abs(2*x-3)$

Enter: $y=7$

Work. The V-shaped graph intersects the horizontal line at (−2, 7) and (5, 7). Both x-coordinates are solutions. A window containing negative inputs is essential.

Answer: $x = -2$ or $x = 5.$

Check. $|2(-2) - 3| = | - 7| = 7$ and $|2(5) - 3| = |7| = 7.$ The two linear cases explain why no other solutions occur.

Avoid the trap. Replacing $|2x - 3| = 7$ with only $2x - 3 = 7$ discards the left branch. The graph is useful precisely because it exposes both cases.

2.07. Intersect a circle with a horizontal line

The circle (x − 2)² + (y + 1)² = 25 intersects y = 3. What is the greater x-coordinate of intersection?

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Recognize. Enter the circle as an implicit equation and the line separately.

Enter: $(x-2)^2+(y+1)^2=25$

Enter: $y=3$

Work. The points of intersection are (−1, 3) and (5, 3). Read the greater horizontal coordinate. A useful view contains $-4 \leq x \leq 8$ and $-7 \leq y \leq 5.$

Answer: 5.

Check. Substitute $y = 3:$ $(x - 2) 2 + 16 = 25,$ so $(x - 2) 2 = 9$ and $x = 2 \pm 3.$

Avoid the trap. The circle's center is (2, −1), not (−2, 1). A graph can look stretched when axis scales differ; use coordinates, not the apparent visual shape, to determine a length.

2.08. Look beyond the initial window

A graph of y = x² and y = 625 shows no intersection in the current view. How many real solutions does x² = 625 have, and what are they?

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Recognize. The output 625 and inputs near $\pm25$ may both lie outside the visible region.

Enter: $y=x^2$

Enter: $y=625$

Work. Set approximately $-30 \leq x \leq 30$ and $-100 \leq y \leq 850.$ The intersections are (−25, 625) and (25, 625). The graph had been incomplete evidence, not evidence of no solutions.

Answer: Two real solutions: −25 and 25.

Check. $252 = 625$ and $(-25) 2 = 625.$ Factoring $x 2 - 625 = (x - 25)(x + 25)$ confirms the complete set.

Avoid the trap. Do not press zoom-out repeatedly without a scale estimate. A one-line estimate of $\sqrt{625}$ tells you where to look.

2.09. Choose different scales for the two axes

Two pricing models are y = 2x + 400 and y = −3x + 900. At what input do they predict the same value?

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Recognize. Both the input and output at intersection are much larger than ordinary default bounds.

Enter: $y=2*x+400$

Enter: $y=-3*x+900$

Work. Try $0 \leq x \leq 200$ and $300 \leq y \leq 950.$ The intersection is (100, 600). The question asks for the input, so report 100.

Answer: 100.

Check. $2(100) + 400 = 600$ and $-3(100) + 900 = 600.$ Equating the models gives $5x = 500.$

Avoid the trap. Expanding only the horizontal axis is not enough if the shared output remains above the screen. The vertical scale is part of a useful graph setup.

2.10. Do not confuse an asymptote with a solution

Find the real solutions of $\frac{6}{x − 1}$ = x.

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Recognize. The original denominator excludes $x = 1.$ The vertical asymptote is not a graph intersection.

Enter: $y=6/(x-1)$

Enter: $y=x$

Work. The curves intersect at (−2, −2) and (3, 3). Both are permitted inputs. The vertical behavior near $x = 1$ does not describe a third root.

Answer: $x = -2$ or $x = 3.$

Check. $6/(-3) = -2$ and $6/2 = 3.$ Multiplying by the nonzero denominator yields $x 2 -x-6 = (x-3)(x+2) = 0,$ establishing completeness.

Avoid the trap. Entering $6/x - 1$ changes the denominator. Also, a steep or nearly vertical piece of a curve is not a solution merely because it lies close to a different graph in a coarse view.

2.11. Graph the original radical equation

Solve $\sqrt{x + 2}$ = x − 4 over the real numbers.

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Recognize. The square root is nonnegative, so any solution must satisfy $x \geq 4.$

Enter: $y=sqrt(x+2)$

Enter: $y=x-4$

Work. The original graphs meet at (7, 3). If you square first, you obtain $x + 2 = (x - 4) 2,$ or $(x - 2)(x - 7) = 0.$ The algebraic candidate $x = 2$ cannot satisfy the original nonnegative-output condition.

Answer: $x = 7.$

Check. $\sqrt{7 + 2}$ At $x = 2,$ the left side is 2 and the right side is −2, so that candidate fails.

Avoid the trap. Squaring can create a different equation with more solutions. A graph of the squared equation alone does not filter those extraneous values.

2.12. Recognize tangency as one intersection

For what value of k does y = x² + 6x + 13 intersect the horizontal line y = k at exactly one point?

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Recognize. A horizontal line meets an upward-opening parabola once when it passes through the vertex.

Enter: $y=x^2+6*x+13$

Work. The graph suggests the vertex (−3, 4). Complete the square: $x 2 + 6x + 13 = (x + 3) 2 + 4.$ Thus $k = 4$ produces one point; $k > 4$ produces two, and $k < 4$ produces none.

Answer: $k = 4.$

Check. At $k = 4,$ the equation is $(x + 3) 2 = 0,$ giving exactly the single distinct input $x = -3.$

Avoid the trap. A moving horizontal slider can suggest the answer but does not establish exact tangency. A line that appears to touch may actually miss or meet twice at a very small scale.

2.13. Interpret the overlap and include the boundary

Find the greatest possible x satisfying 2x + y ≤ 10, y ≥ x + 1, and x ≥ 0.

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Recognize. The feasible region is the overlap of all three inequalities. Non-strict boundaries are included.

Enter: $2*x+y<=10$

Enter: $y>=x+1$

Enter: $x>=0$

Work. The rightmost feasible point occurs where the two sloping boundaries meet: $2x + (x + 1) = 10,$ so $x = 3$ and $y = 4.$ The overlap has vertices (0, 1), (0, 10), and (3, 4).

Answer: 3.

Check. Combining $y \geq x + 1$ with $2x + y \leq 10$ gives $3x + 1 \leq 10,$ hence $x \leq 3.$ The point (3, 4) attains that bound.

Avoid the trap. Points satisfying only one shaded inequality are not necessarily feasible. Shading is an aid; the combined inequalities prove that no larger x works.

2.14. Find every intersection of two nonlinear graphs

How many real ordered pairs satisfy x² + y² = 25 and y = x² − 5?

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Recognize. A circle and a parabola may intersect more than twice. Both graphs need a window that includes their lower and upper portions.

Enter: $x^2+y^2=25$

Enter: $y=x^2-5$

Work. The intersections are (0, −5), (−3, 4), and (3, 4). Use a view such as $-6 \leq x \leq 6$ and $-6 \leq y \leq 6,$ then verify the lower touching point as well as the two upper crossings.

Answer: 3 real ordered pairs.

Check. Substitute $y = x 2 - 5:$ $x 2 + (x 2 - 5) 2 = 25,$ which simplifies to $x 2 (x 2 - 9) = 0.$ Therefore $x = 0,$ $\pm3,$ producing exactly those three points.

Avoid the trap. The lower intersection is a contact, not a crossing. A visual method that counts only crossings undercounts the solutions.

2.15. Graph the unknown parameter as an input

The graph of y = (x − h)² + 2 passes through (5, 11). Find every possible value of h.

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Recognize. The unknown is a parameter. Substitute the known point first, then solve a one-variable equation for that parameter.

Enter: $y=(5-x)^2+2$

Enter: $y=11$

Work. The point condition is $(5 - h) 2 + 2 = 11.$ In the temporary graph setup, the horizontal variable represents h, not the original graph's x. The intersections have horizontal coordinates 2 and 8. Both are possible horizontal shifts.

Answer: $h = 2$ or $h = 8.$

Check. For $h = 2,$ $(5 - 2) 2 + 2 = 11.$ For $h = 8,$ $(5 - 8) 2 + 2 = 11.$ Algebra gives $(5 - h) 2 = 9,$ hence $5 - h = \pm3.$

Avoid the trap. Finding one slider value that makes the original parabola pass through the point does not rule out another. Convert the condition into an equation so that both parameter values can be seen and justified.

The horizontal coordinate is the parameter

Original explanatory graph. The calculator recipe uses x as a temporary stand-in for the unknown h; the mathematical question still asks for h.

Section synthesis: graph, interpret, certify

A graph is a representation of the relationship, not a substitute for the question. Record the full point, apply domain restrictions, and calculate the requested quantity. For exact counts or parameter conditions, explain why the candidates are complete. A different window can reveal more detail, but only valid reasoning can establish completeness.

2.6 Section exit check

Work without the lesson open. Choose a method, show enough reasoning to check the result, and record any tool or interpretation error. These questions include tool-decision drills as well as mathematical problems.

Topic practice

Before checking the solutions

Name one question where a manual method was shorter, one place where calculator setup needed care, and one check that would catch a plausible wrong answer. If you cannot name an independent check, return to the relevant worked example.