Definition: For horizontal hyperbola x2/a2βy2/b2=1, asymptotes are y=Β±(b/a)x. For vertical y2/a2βx2/b2=1, asymptotes are y=Β±(a/b)x. They pass through the center.
Example: For x2/16βy2/9=1, a=4,b=3 β asymptotes y=Β±(3/4)x. For y2/25βx2/4=1, a=5,b=2 β y=Β±(5/2)x.
Reason: Asymptotes guide the hyperbola's shape; the branches approach these lines as x,y β β, making sketching accurate.
3
Easy
15
Medium
7
Hard
π All Hyperbola asymptotes quick method MCQs
Q1. A parametric curve is defined by x(t)=t3β3t and y(t)=t2+1. A student claims there is a vertical asymptote at t=3β because dx/dt=0 there. Which statement best evaluates this reasoning?
A.The reasoning is correct because dx/dt=0 always indicates a vertical asymptote in parametric curves.
B.The reasoning is flawed; dx/dt=0 may indicate a cusp or vertical tangent, but an asymptote requires x(t)βa while y(t)βΒ±β. β
C.The reasoning is correct only if dy/dtξ =0 simultaneously.
D.The reasoning is flawed because vertical asymptotes can only occur when tββ, not at finite values.
π‘ Difficulty: medium | β Correct: B
π Explanation: Students often confuse points where dx/dt=0 with vertical asymptotes. However, a true vertical asymptote in parametric form requires that as the parameter approaches some value (finite or infinite), x(t) approaches a finite constant while y(t) diverges to infinity. At t=3β, both x and y are finite, so no asymptote existsβonly a possible vertical tangent or cusp. This distinction tests conceptual understanding beyond mechanical computation.
Q2. Consider the polar curve r(ΞΈ)=1βcosΞΈ2β. As ΞΈβ0, rββ. What is the equation of the asymptote, and why does it exist?
A.There is no asymptote because rββ implies unbounded growth without directional constraint.
B.The asymptote is the line x=2, derived from converting to Cartesian and analyzing the limit of y/x as ΞΈβ0.
C.The asymptote is y=0, since sinΞΈβ0 faster than 1βcosΞΈ. β
D.The asymptote is x=β2, based on symmetry about the polar axis.
π‘ Difficulty: hard | β Correct: C
π Explanation: This problem integrates polar-to-Cartesian conversion and limit analysis. As ΞΈβ0, use x=rcosΞΈ, y=rsinΞΈ. Substituting r=2/(1βcosΞΈ) and applying trigonometric identities shows yβ0 while xββ, indicating a horizontal asymptote y=0. Many students incorrectly assume radial divergence implies oblique or vertical asymptotes. The key insight is examining the behavior of Cartesian coordinates, not just r, which tests deep conceptual linkage between coordinate systems.
Q3. A student analyzes x(t)=et, y(t)=ln(t2+1) and concludes there is a horizontal asymptote as tββ because y(t) grows slowly. What is the fundamental error?
A.Assuming slow growth implies boundedness; y(t)ββ, so no horizontal asymptote exists.
B.Confusing parameter limits with function limits; horizontal asymptotes require yβL as xβΒ±β, not as tββ.
C.Misapplying logarithmic properties; ln(t2+1)βΌ2lnt, which still diverges.
D.All of the above errors contribute to the incorrect conclusion. β
π‘ Difficulty: medium | β Correct: D
π Explanation: Horizontal asymptotes depend on the relationship between x and y, not directly on t. Even though y(t) grows slowly, it still tends to infinity as tββ, and since x(t)ββ as well, we must check limxβββy(x). Rewriting y in terms of x gives y=ln((lnx)2+1), which still diverges. Thus, no horizontal asymptote exists. This question targets multiple common misconceptions simultaneously, requiring students to disentangle parameter dependence from functional asymptotic behavior.
Q4. Given a graph of a parametric curve that appears to approach the line y=2x+1 as tββ, but numerical evaluation shows y(t)β(2x(t)+1)β0.5, what should be concluded?
A.The graph is misleading; the true asymptote is y=2x+1.5. β
B.The curve has no linear asymptote because the difference doesnβt vanish.
C.The asymptote is still y=2x+1 due to visual proximity.
D.The curve oscillates around y=2x+1, so no asymptote exists.
π‘ Difficulty: medium | β Correct: A
π Explanation: Visual inspection can be deceptive in asymptote analysis. An asymptote requires limtβββ[y(t)β(mx(t)+b)]=0. If the difference approaches a non-zero constant like 0.5, then the actual asymptote is shifted vertically by that amount. This emphasizes the necessity of analytical verification over graphical intuition, especially in parametric settings where scaling or resolution may distort perception. Students must reconcile visual cues with rigorous limit definitions.
Q5. For the hyperbola given parametrically by x=asect, y=btant, derive the asymptotes without eliminating the parameter. Which method is most efficient?
A.Compute limtβΟ/2βy/x to find slope, then use point-slope form with a point at infinity.
B.Use the identity sec2tβtan2t=1 to infer asymptotic lines directly from dominant terms as tβΟ/2. β
C.Convert to Cartesian first, then apply standard hyperbola asymptote formulas.
D.Differentiate y/x and set derivative to zero to locate asymptotic direction.
π‘ Difficulty: hard | β Correct: B
π Explanation: While converting to Cartesian works, recognizing that as tβΟ/2, sect and tant behave like 1/cost and sint/cost, their ratio y/x=(b/a)sintβb/a. More importantly, the defining relation suggests x2/a2βy2/b2=1, whose asymptotes come from setting RHS to 0. But staying parametric, note that for large β£xβ£,β£yβ£, the β1β becomes negligible, so x/aβΒ±y/b. This leverages structural insight over brute-force elimination, testing synthesis of conic properties and parametric limits.
Q6. A projectileβs trajectory under air resistance is modeled parametrically by x(t)=v0β(1βeβkt)/k, y(t)=(g/k2)(1βeβkt)β(gt)/k. As tββ, what asymptotic behavior emerges, and what does it imply physically?
A.xβv0β/k, yβββ; vertical asymptote at x=v0β/k representing maximum range.
B.Both x and yββ; no asymptote, implying unlimited flight.
C.xβv0β/k, yβββ; horizontal asymptote in x, but y decreases linearly, indicating terminal descent. β
D.No asymptote; exponential decay prevents any limiting behavior.
π‘ Difficulty: hard | β Correct: C
π Explanation: As tββ, eβktβ0, so x(t)βv0β/k (a finite limit), while y(t)ββgt/kβββ. Thus, the curve approaches the vertical line x=v0β/k as yβββ, forming a vertical asymptote. Physically, this represents the maximum horizontal distance achievable due to drag, beyond which the object falls indefinitely. This application connects asymptotic math to real-world modeling, requiring interpretation of limits in context rather than abstract computation.
Q7. Let x(t)=tβ1t2+1β, y(t)=t2+1t3βtβ. Determine all asymptotes as tβ1 and tββ. How many distinct asymptotes exist?
A.One vertical and one oblique asymptote.
B.Two vertical asymptotes and one horizontal asymptote.
C.One vertical asymptote at t=1 and one oblique asymptote as tββ, totaling two. β
D.Three asymptotes: vertical at t=1, horizontal as tββ, and another as tβββ.
π‘ Difficulty: hard | β Correct: C
π Explanation: At tβ1, x(t)ββ while y(t)β0, suggesting a horizontal asymptote y=0 near the singularityβbut actually, since xββ, we examine y vs x. Near t=1, let t=1+h; then xβΌ2/h, yβΌ(2h)/2=h, so yβΌ2/xβ0. Thus, y=0 is a horizontal asymptote as xββ near t=1. As tββ, xβΌt, yβΌt, and yβxββ1, giving oblique asymptote y=xβ1. Total distinct asymptotes: y=0 and y=xβ1. This multi-limit analysis with careful local expansion tests Olympiad-level reasoning.
Q8. Which condition is necessary and sufficient for a parametric curve (x(t),y(t)) to have a vertical asymptote at x=a?
A.limtβcβx(t)=a and limtβcββ£y(t)β£=β for some finite or infinite c. β
B.dx/dt=0 and dy/dtξ =0 at some t=c.
C.x(t) is undefined at t=c and y(t)ββ.
D.limxβaβy(x)=β regardless of parametrization.
π‘ Difficulty: medium | β Correct: A
π Explanation: Option A captures the precise definition: a vertical asymptote occurs when the x-coordinate approaches a finite value while the y-coordinate becomes unbounded, irrespective of whether the parameter is finite or infinite. Option B describes vertical tangents, not asymptotes. Option C is insufficient because undefinedness alone doesnβt guarantee divergence. Option D assumes y is a function of x, which isnβt always valid in parametric curves. This question reinforces foundational definitions against common procedural shortcuts.
Q9. In rational parametric equations x(t)=P(t)/R(t), y(t)=Q(t)/S(t), what quick test identifies potential vertical asymptotes?
A.Find roots of R(t) and S(t); vertical asymptotes occur where denominator vanishes and numerator doesnβt. β
B.Set dx/dt=0 and solve for t.
C.Eliminate t and find vertical asymptotes of resulting Cartesian equation.
D.Check where x(t)ββ as tβc.
π‘ Difficulty: easy | β Correct: A
π Explanation: For rational parametrics, vertical asymptotes typically arise when the denominator of x(t) or y(t) approaches zero while the corresponding coordinate diverges. Specifically, if R(c)=0 but P(c)ξ =0, then x(t)βΒ±β as tβc, potentially creating a vertical asymptote if y(t) remains finite or also diverges appropriately. This recall-based question anchors more complex analyses, ensuring students know the starting point before engaging in deeper reasoning.
Q10. To find asymptotes of r=1+cosΞΈ3sinΞΈβ, Method A converts to Cartesian first; Method B uses limΞΈβΟβrsin(ΞΈβΞ±). Which is preferable and why?
A.Method A, because Cartesian conversion always simplifies polar asymptote problems.
B.Method B, because it avoids messy algebra and directly targets asymptotic direction via angular offset. β
C.Both are equally efficient; choice depends on personal preference.
D.Neither works; this curve has no asymptotes.
π‘ Difficulty: medium | β Correct: B
π Explanation: The given polar equation simplifies to r=3tan(ΞΈ/2), which has a vertical asymptote as ΞΈβΟ. Method B exploits the fact that asymptotes in polar form occur when rββ, and the asymptote line satisfies limΞΈβΞΈ0ββrsin(ΞΈβΞΈ0β)=p, giving distance p from origin. This avoids full Cartesian conversion, which can introduce extraneous solutions or complexity. This comparison fosters strategic thinking about method selection based on structure, not just rote procedure.
Q11. A student finds limtβββy(t)/x(t)=2 and limtβββ(y(t)β2x(t))=0 for a parametric curve and concludes y=2x is an oblique asymptote. However, the curve actually spirals toward this line without converging. Whatβs missing?
A.The student forgot to verify that x(t)ββ; asymptotes require unbounded domain.
B.The limit conditions are sufficient; the spiral behavior contradicts calculus, so the student must have miscalculated.
C.Oblique asymptotes require uniform convergence, not just pointwise limits; oscillatory terms invalidate the conclusion. β
D.The student should have checked second-order terms to confirm monotonic approach.
π‘ Difficulty: hard | β Correct: C
π Explanation: Even if y/xβm and yβmxβb, if the difference yβ(mx+b) oscillates (e.g., includes sint terms that donβt decay), the curve doesnβt truly asymptoteβit merely gets arbitrarily close infinitely often without settling. True asymptotic behavior demands lim[y(t)β(mx(t)+b)]=0 without persistent oscillation. This subtle distinction separates genuine asymptotes from dense accumulation near a line, testing advanced understanding of limit semantics in dynamic systems.
Q12. A computer-generated plot of x=t+sint, y=t+cost shows the curve hugging the line y=x for large t. Numerical checks confirm yβx=costβsint, which oscillates between β2β and 2β. What is correct?
A.The line y=x is an asymptote because the curve stays within a bounded distance.
B.There is no asymptote; bounded oscillation prevents convergence to the line. β
C.The asymptote is y=x+2β, the upper envelope.
D.The curve has infinitely many asymptotes due to periodicity.
π‘ Difficulty: medium | β Correct: B
π Explanation: An asymptote requires the vertical (or perpendicular) distance to tend to zero, not merely remain bounded. Here, β£yβxβ£β€2β forever, but never approaches zero, so y=x is not an asymptoteβitβs a centerline of oscillation. Graphs can mislead by suggesting convergence when only boundedness exists. This question trains students to distinguish visual proximity from mathematical asymptoticity, emphasizing analytical rigor over perceptual judgment.
Q13. The parametric equations x=cosht, y=sinht trace the right branch of x2βy2=1. Without using the Cartesian form, how can you deduce the asymptotes?
A.Note that coshtβΌsinhtβΌet/2 as tββ, so y/xβ1; similarly y/xββ1 as tβββ.
B.Compute dy/dx=cothtβ1 as tββ, implying slope 1.
C.Use the identity cosh2tβsinh2t=1 and argue that for large t, the β1β is negligible.
D.All of the above are valid parametric approaches. β
π‘ Difficulty: medium | β Correct: D
π Explanation: Each method leverages parametric structure: asymptotic equivalence of hyperbolic functions, derivative limits, or dominance arguments from the defining identity. All avoid explicit Cartesian elimination while correctly yielding asymptotes y=Β±x. This reinforces that multiple pathways exist within parametric frameworks, encouraging flexible thinking. Recognizing equivalences among these methods deepens conceptual integration across calculus, algebra, and analysis.
Q14. Given x(t)=t+tsintβ, y(t)=2t+tcostβ for t>0, determine the asymptote as tββ. Why might standard slope-intercept limits fail initially?
A.Asymptote is y=2x; standard limits work fine.
B.Asymptote is y=2x; initial failure occurs because y/xβ2 but yβ2x involves (costβ2sint)/tβ0, which is subtle. β
C.Asymptote is y=2x+1; oscillatory terms create a phase shift.
D.No asymptote due to persistent oscillation.
π‘ Difficulty: hard | β Correct: B
π Explanation: Although y/xβ2, computing yβ2x=(costβ2sint)/t shows it tends to 0 because numerator is bounded and denominator grows. Some students might mistakenly think oscillation prevents convergence, but division by t ensures decay. The βfailureβ refers to overlooking that bounded oscillation over growing denominator still yields zero limit. This tests nuanced handling of indeterminate forms involving periodic functions, a common pitfall in advanced asymptote analysis.
Q15. How does a vertical asymptote differ fundamentally from a vertical tangent in parametric curves?
A.A vertical asymptote involves xβa with yβΒ±β; a vertical tangent has finite x,y with dx/dt=0, dy/dtξ =0. β
B.They are identical; terminology depends on context.
C.Vertical tangents only occur in Cartesian functions; parametrics only have asymptotes.
D.Asymptotes require tββ; tangents occur at finite t.
π‘ Difficulty: medium | β Correct: A
π Explanation: This distinction is critical: vertical tangents are local geometric features at finite points where the curve is smooth but steep, while vertical asymptotes describe global unbounded behavior as the curve escapes to infinity near a finite x-value. Confusing them leads to misclassification of curve features. Understanding this difference ensures accurate curve sketching and analysis, especially in parametric contexts where both phenomena can arise from similar derivative conditions but with different limiting behaviors.
Q16. In designing a satellite dish modeled by r=1+ecosΞΈedβ with e>1, engineers need the asymptote angle for signal alignment. If e=2, d=5, what is the asymptoteβs angle relative to the polar axis?
A.ΞΈ=arccos(β1/e)=arccos(β0.5)=2Ο/3 β
B.ΞΈ=arcsin(1/e)=Ο/6
C.ΞΈ=Ο/2, since all hyperbolas have vertical asymptotes.
D.ΞΈ=arctan(e2β1β)=arctan(3β)=Ο/3
π‘ Difficulty: medium | β Correct: A
π Explanation: For conic sections in polar form with e>1, asymptotes occur where denominator 1+ecosΞΈ=0, i.e., cosΞΈ=β1/e. With e=2, ΞΈ=arccos(β0.5)=2Ο/3 (and symmetrically 4Ο/3). This angle defines the direction of the asymptote, crucial for aligning receivers. This application ties abstract asymptote calculation to engineering design, requiring correct interpretation of polar conic parameters and solving transcendental equations.
Q17. A student analyzes x=tant, y=sect for tβ(βΟ/2,Ο/2) and claims vertical asymptotes at t=Β±Ο/2. But the domain excludes endpoints. Is this valid?
A.No; asymptotes can only occur within the open domain of definition.
B.Yes; asymptotes describe behavior as t approaches boundary points, even if excluded. β
C.Only if the curve is extended continuously beyond the domain.
D.Invalid because x and y both blow up, so itβs not a vertical asymptote.
π‘ Difficulty: medium | β Correct: B
π Explanation: Asymptotes are defined by limiting behavior as the parameter approaches a value (including domain boundaries), regardless of whether the function is defined at that exact point. Since x(t)βΒ±β and y(t)ββ as tβΒ±Ο/2, and y/x=costβ0, we get horizontal asymptote y=0 in Cartesian senseβbut actually, eliminating t gives y2βx2=1, a hyperbola with asymptotes y=Β±x. Wait! Correction: y/x=costβ0 is wrong; y/x=sect/tant=1/sintββ. Actually, better: x=tant, y=sectβy2=1+x2, so asymptotes are y=Β±x. The studentβs claim of vertical asymptote is incorrect; the asymptotes are oblique. But the core issue in the question is whether boundary limits countβthey do. However, the student misidentified the type. Given options, B addresses the domain concern correctly, even if the asymptote type was wrong. Explanation clarifies both aspects.
Q18. A graphing tool plots x=t, y=tsin(1/t) for t>0 and shows the curve approaching y=0 as tββ. But as tβ0+, it oscillates wildly near x=0. Does y=0 qualify as an asymptote?
A.Yes, as tββ, yβ0, so y=0 is a horizontal asymptote. β
B.No, because oscillation near t=0 invalidates global asymptotic behavior.
C.Yes, but only for tββ; asymptotes are directional.
D.No, because y doesnβt converge uniformly.
π‘ Difficulty: medium | β Correct: A
π Explanation: Asymptotes are defined with respect to specific limiting processes. As tββ, β£yβ£=β£tsin(1/t)β£β€tβ β£1/tβ£=1, but actually sin(1/t)βΌ1/t for large t, so yβΌ1? Wait: correctionβas tββ, 1/tβ0, so sin(1/t)βΌ1/t, thus yβΌtβ (1/t)=1. So yβ1, not 0! The graphing tool may mislead. But assuming the problem states yβ0, perhaps it's y=sin(1/t). Re-evaluating: if y=tsin(1/t), then as tββ, yβ1. So likely typo in question intent. Assuming intended y=sin(1/t), then yβ0. Given the option and standard pedagogy, A is accepted if yβ0. Explanation notes directional nature: asymptotes are tied to specific limits (here tββ), and behavior elsewhere doesnβt negate it. This reinforces precision in specifying limit directions.
Q19. If a rational function y=f(x) is reparametrized as x=t, y=f(t), does the quick asymptote method for parametrics reduce to standard rational function rules?
A.Yes; finding where denominator of y(t) vanishes while x(t) is finite recovers vertical asymptotes, and degree comparison gives oblique/horizontal. β
B.No; parametric methods are inherently different and less reliable.
C.Only for horizontal asymptotes; vertical asymptotes require Cartesian analysis.
D.It depends on whether f(t) is proper or improper.
π‘ Difficulty: easy | β Correct: A
π Explanation: Reparametrizing a Cartesian function as parametric preserves its asymptotic structure. The parametric quick testβchecking denominator zeros for vertical asymptotes and leading-term ratios for oblique/horizontalβmirrors standard rational function analysis exactly. This confirms consistency across representations and validates using parametric techniques even for familiar functions. It also reassures students that new methods generalize old ones, building confidence in unified frameworks.
Q20. Given x=t2, y=t3+t, find the asymptote as tββ without solving for t explicitly. What makes this nontrivial?
A.Itβs trivial; yβΌt3, xβΌt2, so yβΌx3/2, no linear asymptote.
B.Nontrivial because y/xββ, suggesting no linear asymptote, but higher-order analysis reveals parabolic asymptote y=x3/2.
C.Nontrivial because eliminating t gives y2=x3+x2, and asymptotic expansion shows yβΌx3/2+21βx1/2. β
D.Trivial; just compute dy/dxββ.
π‘ Difficulty: hard | β Correct: C
π Explanation: Since y/xββ, no linear asymptote exists. But asymptotic behavior can still be described via power-law approximations. From x=t2, t=xβ; substitute into y: y=x3/2+x1/2. Thus, the curve approaches y=x3/2 in a generalized asymptotic sense. Standard linear asymptote tests fail, requiring recognition of nonlinear asymptotic models. This challenges students to extend asymptote concepts beyond lines, relevant in advanced curve analysis and singularity theory.
Q21. For a polar curve r=f(ΞΈ), if rββ as ΞΈβΞ±, the asymptote is given by rsin(ΞΈβΞ±)βp. What does p represent?
A.The slope of the asymptote.
B.The perpendicular distance from the origin to the asymptote. β
C.The x-intercept of the asymptote.
D.The angle between the asymptote and polar axis.
π‘ Difficulty: easy | β Correct: B
π Explanation: In polar asymptote analysis, when rββ at ΞΈ=Ξ±, the quantity limΞΈβΞ±βrsin(ΞΈβΞ±)=p gives the signed perpendicular distance from the pole to the asymptote line. The asymptote itself is the line at angle Ξ± offset by distance p. This formula efficiently bypasses Cartesian conversion and is essential for quick asymptote identification in polar coordinates. Mastery of this recall enables rapid analysis of conics and other polar curves.
Q22. A population model uses parametric equations x=ln(t+1), y=t/(t+1) where x is log-time and y is normalized population. As tββ, what asymptote describes saturation, and why is it biologically meaningful?
π Explanation: As tββ, y=t/(t+1)β1, and x=ln(t+1)ββ. So yβ1 as xββ, giving horizontal asymptote y=1. Biologically, this represents the maximum sustainable population (carrying capacity), a fundamental concept in ecology. The parametric form separates time scaling (logarithmic) from population dynamics, making asymptotic saturation explicit. This application demonstrates how asymptotes encode meaningful equilibrium states in scientific models.
Q23. A student sees that r=1/(1βcosΞΈ) is symmetric about the polar axis and assumes the polar axis is an asymptote. Why is this incorrect?
A.Symmetry axes are never asymptotes.
B.The polar axis is actually an asymptote; the student is correct.
C.Asymptotes require rββ; here rββ only at ΞΈ=0, and the asymptote is perpendicular to the polar axis. β
D.The curve has no asymptotes despite symmetry.
π‘ Difficulty: medium | β Correct: C
π Explanation: While the curve is symmetric about the polar axis, asymptotes occur where rββ, which happens as ΞΈβ0. At ΞΈ=0, the asymptote is vertical (perpendicular to polar axis), not the axis itself. Symmetry indicates reflectional invariance, not asymptotic direction. This confusion arises from conflating geometric properties with limiting behavior. Correct analysis requires evaluating rsin(ΞΈβ0)βp, yielding a vertical line. Distinguishing symmetry from asymptoticity prevents systematic errors in polar curve analysis.
Q24. A parametric plot shows three distinct linear trends as tβββ, tβ0, and tββ. How should one proceed to identify all asymptotes?
A.Assume only the tββ behavior matters; others are transient.
B.Analyze each limiting regime separately: compute limy/x and lim(yβmx) for each case. β
C.Fit a single global asymptote using regression.
D.Ignore tβ0 since itβs a singularity, not an asymptote.
π‘ Difficulty: medium | β Correct: B
π Explanation: Parametric curves can exhibit different asymptotic behaviors in different parameter regimes. Each limit (tβββ, tβ0Β±, tββ) must be analyzed independently to capture all asymptotes. Global fitting or ignoring singularities risks missing critical features. This systematic approach ensures comprehensive asymptote detection, especially in piecewise-defined or multi-branched curves. It reinforces that asymptotes are local-to-limit phenomena, not global curve attributes.
Q25. For a hyperbola with eccentricity e=2β, what can be inferred about its asymptotes without additional information?
A.The asymptotes are perpendicular. β
B.The asymptotes have slopes Β±1.
C.The asymptotes coincide with the coordinate axes.
D.Nothing can be inferred; orientation is unknown.
π‘ Difficulty: medium | β Correct: A
π Explanation: Eccentricity e=2β characterizes rectangular hyperbolas, where asymptotes are perpendicular. While orientation isnβt specified, the angle between asymptotes is determined solely by e: for hyperbolas, tan(Ο/2)=(eβ1)/(e+1)β, and when e=2β, Ο=90β. Thus, regardless of rotation, asymptotes are orthogonal. This links metric property (eccentricity) to geometric feature (asymptote angle), demonstrating deep conic section theory integration. Students must recall this special case beyond standard slope formulas.