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PSAT/NMSQT Math / LEVEL 2 · DIFFICULTY 3/5

Advanced Math: Expressions & Functions

Connect equivalent expressions, nonlinear solutions and function models.

6 stages · 48 practice problems · two 12-question assessment forms

Choose a node to explore its lesson.

  1. 1MINI QUEST PSAT 2.1Rational Exponents & RadicalsRead the lesson
  2. 2MINI QUEST PSAT 2.2Quadratic Formula & Completing the SquareRead the lesson
  3. 3MINI QUEST PSAT 2.3Rational Equations & Domain RestrictionsRead the lesson
  4. 4MINI QUEST PSAT 2.4Radical Equations & Extraneous RootsRead the lesson
  5. 5MINI QUEST PSAT 2.5Lines & Parabolas: Nonlinear SystemsRead the lesson
  6. 6MINI QUEST PSAT 2.6Exponential Growth, Decay & TablesRead the lesson
  7. LEVEL CHECKCastle challenge

    12 questions across this level.

    Sign in for the level check →

STAGE PSAT 2.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Rational Exponents & Radicals

The idea to remember

For positive a, amn\displaystyle a^{\frac{m}{n}}anm​ means the nth root of a to power m. Even real roots require nonnegative radicands.

Remember with 2 flashcards →

Goal: Rewrite roots as fractional powers and simplify equivalent expressions with valid domain restrictions.

Before you begin: Read the problem carefully and use the linked foundation lessons when a term or operation is unfamiliar.

SAT extension: rational exponents and radical rewriting are enrichment beyond the PSAT equivalent-expression requirements.

Make sense of the idea

For positive a, amn\displaystyle a^{\frac{m}{n}}anm​ means the nth root of a, raised to power m. The denominator names the root; the numerator names the power.

Choose a method

For a>0\displaystyle a > 0a>0, a−r=1ar\displaystyle a^{-r} = \frac{1}{a^{r}}a−r=ar1​. When multiplying powers with the same base, add exponents. Do not add the bases.

Check and interpret

Over real numbers, even roots require nonnegative radicands. The principal square root is nonnegative: x2=∣x∣\displaystyle \sqrt{x^{2}} = \left|x\right|x2​=∣x∣, not always x. Keep denominator restrictions when simplifying rational expressions.

WORKED EXAMPLE 1

Evaluate 6423\displaystyle 64^{\frac{2}{3}}6432​.

  1. The cube root of 64 is 4.
  2. 6423=42=16\displaystyle 64^{\frac{2}{3}} = 4^{2} = 166432​=42=16.

16

WORKED EXAMPLE 2

A positive-input model contains x16×x56\displaystyle x^{\frac{1}{6}} \times x^{\frac{5}{6}}x61​×x65​. What is the exponent of x after combining the factors?

  1. 16+56=66\displaystyle \frac{1}{6} + \frac{5}{6} = \frac{6}{6}61​+65​=66​.
  2. The combined power is x1\displaystyle x^{1}x1, so the exponent is 1.

1

Common pitfalls

Possible mix-up: A negative exponent makes the result negative.

It takes a reciprocal. A positive base remains positive.

Possible mix-up: x2\displaystyle \sqrt{x^{2}}x2​ always equals x.

It equals ∣x∣\displaystyle \left|x\right|∣x∣; test a negative value of x.

Explain it to yourself

In rational exponents & radicals, what information determines your method, and how can you check whether your result fits the original conditions?

Discuss together

Ask your learner to explain one representation and check the conditions in rational exponents & radicals. Compare independent answers with hinted attempts and application problems. A completed lesson is not a PSAT score prediction.

Preview the eight practice prompts
  1. Problem 1

    Evaluate 823\displaystyle 8^{\frac{2}{3}}832​.

    Practise this exact problem →
  2. Problem 2

    Evaluate 4−12\displaystyle 4^{\frac{-1}{2}}42−1​.

    Practise this exact problem →
  3. Problem 3

    Evaluate x2\displaystyle \sqrt{x^{2}}x2​ when x=−3\displaystyle x = -3x=−3.

    Practise this exact problem →
  4. Problem 4

    For x>0\displaystyle x > 0x>0, an expression contains x13×x23\displaystyle x^{\frac{1}{3}} \times x^{\frac{2}{3}}x31​×x32​. What is the exponent of x after combining the factors?

    Practise this exact problem →
  5. Problem 5New context

    A scaling rule requires evaluating 2723\displaystyle 27^{\frac{2}{3}}2732​.

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  6. Problem 6New context

    A model has multiplier 9−12\displaystyle 9^{\frac{-1}{2}}92−1​.

    Practise this exact problem →
  7. Problem 7New context

    A distance calculation contains x2\displaystyle \sqrt{x^{2}}x2​ when x=−4\displaystyle x = -4x=−4.

    Practise this exact problem →
  8. Problem 8New context

    A positive-input model contains x14×x34\displaystyle x^{\frac{1}{4}} \times x^{\frac{3}{4}}x41​×x43​. What is the exponent of x after combining the factors?

    Practise this exact problem →
Open stage PSAT 2.1 in the student workspace →

STAGE PSAT 2.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Quadratic Formula & Completing the Square

The idea to remember

The quadratic formula solves ax2+bx+c=0\displaystyle ax^{2} + bx + c = 0ax2+bx+c=0 when a≠0\displaystyle a \ne 0a=0. Completing the square reveals the vertex form.

Remember with 2 flashcards →

Goal: Choose a quadratic-solving method, retain both roots, and interpret the discriminant.

Before you begin: Read the problem carefully and use the linked foundation lessons when a term or operation is unfamiliar.

Parent quadratic: y=x2\displaystyle y = x^{2}y=x2. Explore the parabola, then use the worked examples below to connect it to solutions and intersections.

AN EQUATION IS A RULE FOR MAKING POINTS

Change the rule. Watch the graph.

Pick an x value. The rule tells you its y value. A point (x, y) shows both numbers together. Drag across the graph or use the point slider. This is an adjustable learning example; use the coefficients in your own practice question when solving it.

Parent equation: y=x2\displaystyle y = x^{2}y=x2

Start with y=x2\displaystyle y = x^{2}y=x2. Complete the square: y=1 (x−(1))2+(−4)\displaystyle y = 1\, \left(x - \left(1\right)\right)^{2} + \left(-4\right)y=1(x−(1))2+(−4). The new vertex is (1, -4). The first coordinate gives the horizontal shift and the second gives the vertical shift. Multiply vertical distances from the vertex by 1.

Transform the parent graph
0%

Teal moving graph: y=1 x2+(0) x+(0)\displaystyle y = 1\, x^{2} + \left(0\right)\, x + \left(0\right)y=1x2+(0)x+(0)

Blue: y=1 x2−2 x−3\displaystyle y = 1\, x^{2} - 2\, x - 3y=1x2−2x−3

y = 1x² − 2x − 3; horizontal x-axis and vertical y-axis-10-10-8-8-6-6-4-4-2-2224466881010xy0

Moving point: (-4, 16)

Teal shows the current transformation. Blue is the target. At 100%\displaystyle 100\%100% they coincide; intermediate graphs are learning frames, not additional solutions.

Dashed gray: parent equation. Blue: current equation. View: −10 to 10 on both axes. The graph continues beyond this window. Purple marks your selected point.

Curve coefficient a: 1
Coefficient b: -2
Vertical intercept c: -3
Choose the point’s x: 1
Point (1, -4)
1×(1)2+(−2)×(1)+(−3)=−4\displaystyle 1 \times \left(1\right)^{2} + \left(-2\right) \times \left(1\right) + \left(-3\right) = -41×(1)2+(−2)×(1)+(−3)=−4
Why does it bend?

The x2\displaystyle x^{2}x2 term makes the change in y vary with x. This parabola opens upward because a is positive.

Three points made by the blue rule
xSubstitute into the ruley
-11×(−1)2+(−2)×(−1)+(−3)\displaystyle 1 \times \left(-1\right)^{2} + \left(-2\right) \times \left(-1\right) + \left(-3\right)1×(−1)2+(−2)×(−1)+(−3)0
01×(0)2+(−2)×(0)+(−3)\displaystyle 1 \times \left(0\right)^{2} + \left(-2\right) \times \left(0\right) + \left(-3\right)1×(0)2+(−2)×(0)+(−3)-3
11×(1)2+(−2)×(1)+(−3)\displaystyle 1 \times \left(1\right)^{2} + \left(-2\right) \times \left(1\right) + \left(-3\right)1×(1)2+(−2)×(1)+(−3)-4

Make sense of the idea

Write ax2+bx+c=0\displaystyle ax^{2} + bx + c = 0ax2+bx+c=0 with a≠0\displaystyle a \ne 0a=0. The quadratic formula is x=−b±b2−4 ac2 a\displaystyle x = \frac{-b \pm \sqrt{b^{2} - 4\, ac}}{2\, a}x=2a−b±b2−4ac​​. The denominator 2a divides the entire numerator.

Choose a method

To complete the square in x2+bx\displaystyle x^{2} + bxx2+bx, add (b2)2\displaystyle \left(\frac{b}{2}\right)^{2}(2b​)2. Add it to both sides. For example, x2+6 x=7\displaystyle x^{2} + 6\, x = 7x2+6x=7 becomes (x+3)2=16\displaystyle \left(x + 3\right)^{2} = 16(x+3)2=16, so x=1\displaystyle x = 1x=1 or −7.

Check and interpret

The discriminant D=b2−4 ac\displaystyle D = b^{2} - 4\, acD=b2−4ac counts real roots: D>0\displaystyle D > 0D>0 gives two, D=0\displaystyle D = 0D=0 gives one repeated root, D<0\displaystyle D < 0D<0 gives none. A context can rule out a negative length without deleting it from the algebraic equation. Parent equation: y=x2\displaystyle y = x^{2}y=x2.

WORKED EXAMPLE 1

Solve 2 x2−9 x+4=0\displaystyle 2\, x^{2} - 9\, x + 4 = 02x2−9x+4=0. Use the quadratic formula to find the larger real root.

  1. D=(−9)2−4×2×4=49\displaystyle D = \left(-9\right)^{2} - 4 \times 2 \times 4 = 49D=(−9)2−4×2×4=49, so D=7\displaystyle \sqrt{D} = 7D​=7.
  2. x=9±74\displaystyle x = \frac{9 \pm 7}{4}x=49±7​ gives 4 and 12\displaystyle \frac{1}{2}21​. The larger root is 4.

4

WORKED EXAMPLE 2

A design equation is x2−12 x+c=0\displaystyle x^{2} - 12\, x + c = 0x2−12x+c=0. Find c so that it has exactly one real solution.

  1. (−12)2−4 c=0\displaystyle \left(-12\right)^{2} - 4\, c = 0(−12)2−4c=0.
  2. c=1444=36\displaystyle c = \frac{144}{4} = 36c=4144​=36; then (x−6)2=0\displaystyle \left(x - 6\right)^{2} = 0(x−6)2=0.

36

Common pitfalls

Possible mix-up: Taking a square root gives only the positive branch.

An equation z2=c\displaystyle z^{2} = cz2=c with c>0\displaystyle c > 0c>0 has z = ±c\displaystyle \sqrt{c}c​.

Possible mix-up: Only c is divided by 2a.

In the quadratic formula the entire numerator is divided by 2a.

Explain it to yourself

In quadratic formula & completing the square, what information determines your method, and how can you check whether your result fits the original conditions?

Discuss together

Ask your learner to explain one representation and check the conditions in quadratic formula & completing the square. Compare independent answers with hinted attempts and application problems. A completed lesson is not a PSAT score prediction.

Preview the eight practice prompts
  1. Problem 1

    Solve 2 x2−5 x+2=0\displaystyle 2\, x^{2} - 5\, x + 2 = 02x2−5x+2=0. Use the quadratic formula to find the larger real root.

    Practise this exact problem →
  2. Problem 2

    Complete the square to solve x2+4 x=12\displaystyle x^{2} + 4\, x = 12x2+4x=12. Find the larger root.

    Practise this exact problem →
  3. Problem 3

    How many real roots does x2+4 x+5=0\displaystyle x^{2} + 4\, x + 5 = 0x2+4x+5=0 have? Enter the number of real roots.

    Practise this exact problem →
  4. Problem 4

    The equation is x2−6 x+c=0\displaystyle x^{2} - 6\, x + c = 0x2−6x+c=0. Find c so that it has exactly one real solution.

    Practise this exact problem →
  5. Problem 5New context

    A model reaches zero when 2 x2−7 x+3=0\displaystyle 2\, x^{2} - 7\, x + 3 = 02x2−7x+3=0. Use the quadratic formula to find the larger real root.

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  6. Problem 6New context

    A trajectory equation reduces to x2+6 x=16\displaystyle x^{2} + 6\, x = 16x2+6x=16. Find the larger root.

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  7. Problem 7New context

    A height model would reach zero at a real root of x2+6 x+10=0\displaystyle x^{2} + 6\, x + 10 = 0x2+6x+10=0? Enter the number of real roots.

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  8. Problem 8New context

    A design equation is x2−8 x+c=0\displaystyle x^{2} - 8\, x + c = 0x2−8x+c=0. Find c so that it has exactly one real solution.

    Practise this exact problem →
Open stage PSAT 2.2 in the student workspace →

STAGE PSAT 2.3 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Rational Equations & Domain Restrictions

The idea to remember

Exclude zero denominators before clearing them. Check every candidate in the original equation.

Remember with 2 flashcards →

Goal: Solve equations with variable denominators and reject values outside the original domain.

Before you begin: Read the problem carefully and use the linked foundation lessons when a term or operation is unfamiliar.

Make sense of the idea

First list every value that makes an original denominator zero. These values are forbidden even if a factor later cancels.

Choose a method

Multiply both sides by a common denominator only on the allowed domain. Solve the resulting equation, then substitute every candidate into the original equation.

Check and interpret

An equation can reduce to a contradiction, an identity with excluded points, or a finite set of valid roots. A canceled factor does not erase an original hole in the domain.

WORKED EXAMPLE 1

Solve 18x−4=3\displaystyle \frac{18}{x - 4} = 3x−418​=3. Find x.

  1. 18=3 (x−4)\displaystyle 18 = 3\, \left(x - 4\right)18=3(x−4).
  2. x=10\displaystyle x = 10x=10, whose denominator is 6, not zero.

10

WORKED EXAMPLE 2

A simplified sensor model is x2−49x+7\displaystyle \frac{x^{2} - 49}{x + 7}x+7x2−49​. Which value of x remains excluded after simplification?

  1. x+7=0\displaystyle x + 7 = 0x+7=0 gives x=−7\displaystyle x = -7x=−7.
  2. Although the expression simplifies to x−7\displaystyle x - 7x−7, the value −7 remains excluded.

-7

Common pitfalls

Possible mix-up: A canceled denominator can safely be zero.

The original expression is still undefined there.

Possible mix-up: Every root after clearing denominators is valid.

Check the original equation and excluded values.

Explain it to yourself

In rational equations & domain restrictions, what information determines your method, and how can you check whether your result fits the original conditions?

Discuss together

Ask your learner to explain one representation and check the conditions in rational equations & domain restrictions. Compare independent answers with hinted attempts and application problems. A completed lesson is not a PSAT score prediction.

Preview the eight practice prompts
  1. Problem 1

    Solve 12x−2=3\displaystyle \frac{12}{x - 2} = 3x−212​=3. Find x.

    Practise this exact problem →
  2. Problem 2

    Consider x2−4x−2=4\displaystyle \frac{x^{2} - 4}{x - 2} = 4x−2x2−4​=4. How many real solutions does the ORIGINAL equation have?

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  3. Problem 3

    An equation uses 1x+12=710\displaystyle \frac{1}{x} + \frac{1}{2} = \frac{7}{10}x1​+21​=107​. Solve for x.

    Practise this exact problem →
  4. Problem 4

    The expression is x2−16x+4\displaystyle \frac{x^{2} - 16}{x + 4}x+4x2−16​. Which value of x remains excluded after simplification?

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  5. Problem 5New context

    A rate model requires 15x−3=3\displaystyle \frac{15}{x - 3} = 3x−315​=3. Find x.

    Practise this exact problem →
  6. Problem 6New context

    A calibration equality is x2−9x−3=6\displaystyle \frac{x^{2} - 9}{x - 3} = 6x−3x2−9​=6. How many real solutions does the ORIGINAL equation have?

    Practise this exact problem →
  7. Problem 7New context

    A work-rate equation uses 1x+13=918\displaystyle \frac{1}{x} + \frac{1}{3} = \frac{9}{18}x1​+31​=189​. Solve for x.

    Practise this exact problem →
  8. Problem 8New context

    A simplified sensor model is x2−25x+5\displaystyle \frac{x^{2} - 25}{x + 5}x+5x2−25​. Which value of x remains excluded after simplification?

    Practise this exact problem →
Open stage PSAT 2.3 in the student workspace →

STAGE PSAT 2.4 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Radical Equations & Extraneous Roots

The idea to remember

Squaring may introduce extraneous solutions because opposite signs have the same square. Check the original equation.

Remember with 2 flashcards →

Goal: Use the sign and domain of a square root before squaring, then check each candidate in the original equation.

Before you begin: Read the problem carefully and use the linked foundation lessons when a term or operation is unfamiliar.

PARENT EQUATION

Square-root parent: y=x\displaystyle y = \sqrt{x}y=x​

Starts at (0, 0), curves right and upward, and has real domain x≥0\displaystyle x \ge 0x≥0.

How does the question’s equation change this graph?

For y=a x−h+k\displaystyle y = a\, \sqrt{x - h} + ky=ax−h​+k, the endpoint is (h, k), the domain is x≥h\displaystyle x \ge hx≥h, and negative a reflects vertically.

Check: A square root means the nonnegative principal root. Squaring an equation can introduce answers that fail the original equation.

First identify this family. Then read the coefficients and shifts in your actual equation; this reference does not replace the question’s given values.

Make sense of the idea

The principal square root u\displaystyle \sqrt{u}u​ is nonnegative and is real only if u≥0\displaystyle u \ge 0u≥0. In u=v\displaystyle \sqrt{u} = vu​=v, v must also be nonnegative.

Choose a method

Isolate the radical before squaring. Squaring can lose sign information, so it can introduce candidates that never satisfied the original equation.

Check and interpret

Substitute every candidate back into the unsquared equation. If the left side is a nonnegative root while the right side is negative, reject that candidate.

WORKED EXAMPLE 1

Solve x+4=5\displaystyle \sqrt{x + 4} = 5x+4​=5. Find x.

  1. x+4=25\displaystyle x + 4 = 25x+4=25.
  2. x=21\displaystyle x = 21x=21; substitution gives 25=5\displaystyle \sqrt{25} = 525​=5.

21

WORKED EXAMPLE 2

A side-length condition is 2 x+6+1=8\displaystyle \sqrt{2\, x + 6} + 1 = 82x+6​+1=8. Find x.

  1. 2 x+6=7\displaystyle \sqrt{2\, x + 6} = 72x+6​=7, so 2 x+6=49\displaystyle 2\, x + 6 = 492x+6=49.
  2. x=49−62=432\displaystyle x = \frac{49 - 6}{2} = \frac{43}{2}x=249−6​=243​. Substitution checks both the radicand and sign.

432\displaystyle \frac{43}{2}243​

Common pitfalls

Possible mix-up: Squaring preserves all information.

For instance 22=(−2)2\displaystyle 2^{2} = \left(-2\right)^{2}22=(−2)2 but 2≠−2\displaystyle 2 \ne -22=−2.

Possible mix-up: A negative right side can equal a principal square root.

A principal square root is nonnegative.

Explain it to yourself

In radical equations & extraneous roots, what information determines your method, and how can you check whether your result fits the original conditions?

Discuss together

Ask your learner to explain one representation and check the conditions in radical equations & extraneous roots. Compare independent answers with hinted attempts and application problems. A completed lesson is not a PSAT score prediction.

Preview the eight practice prompts
  1. Problem 1

    Solve x+2=3\displaystyle \sqrt{x + 2} = 3x+2​=3. Find x.

    Practise this exact problem →
  2. Problem 2

    How many real solutions satisfy x+2=−3\displaystyle \sqrt{x + 2} = -3x+2​=−3? Enter the number of solutions.

    Practise this exact problem →
  3. Problem 3

    Solve x−0=x−2\displaystyle \sqrt{x - 0} = x - 2x−0​=x−2. Find the valid real solution.

    Practise this exact problem →
  4. Problem 4

    Solve 2 x+3+1=5\displaystyle \sqrt{2\, x + 3} + 1 = 52x+3​+1=5. Find x.

    Practise this exact problem →
  5. Problem 5New context

    A distance model requires x+3=4\displaystyle \sqrt{x + 3} = 4x+3​=4. Find x.

    Practise this exact problem →
  6. Problem 6New context

    A sensor equation is x+3=−4\displaystyle \sqrt{x + 3} = -4x+3​=−4? Enter the number of solutions.

    Practise this exact problem →
  7. Problem 7New context

    A length equation reduces to x−1=x−3\displaystyle \sqrt{x - 1} = x - 3x−1​=x−3. Find the valid real solution.

    Practise this exact problem →
  8. Problem 8New context

    A side-length condition is 2 x+4+1=6\displaystyle \sqrt{2\, x + 4} + 1 = 62x+4​+1=6. Find x.

    Practise this exact problem →
Open stage PSAT 2.4 in the student workspace →

STAGE PSAT 2.5 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Lines & Parabolas: Nonlinear Systems

The idea to remember

Substitute the line into the nonlinear equation. Keep all resulting intersections that satisfy both originals.

Remember with 2 flashcards →

Goal: Find intersections by substitution and distinguish two, one and no real intersections.

Before you begin: Read the problem carefully and use the linked foundation lessons when a term or operation is unfamiliar.

Two rules. Shared points.

Parent parabola: y=x2\displaystyle y = x^{2}y=x2. The horizontal line y=c\displaystyle y = cy=c is also a linear equation. A solution must lie on both graphs.

-3-2-11232468xy

2 real intersections. (-2, 4) and (2, 4)

Set the outputs equal: x2=4\displaystyle x^{2} = 4x2=4. Both the positive and negative square roots work.

Make sense of the idea

Parent equations: y=x\displaystyle y = xy=x for lines and y=x2\displaystyle y = x^{2}y=x2 for parabolas. An intersection must have the same x and y in both equations.

Choose a method

If both equations give y, set their right sides equal. Rearrange into a quadratic and solve. Substitute each x into either original equation to recover y.

Check and interpret

The quadratic discriminant tells you whether the line crosses the parabola twice, touches once, or misses it. The number of intersections is not automatically two.

WORKED EXAMPLE 1

Find intersections of y=x2\displaystyle y = x^{2}y=x2 and y=10 x−24\displaystyle y = 10\, x - 24y=10x−24. What is the smaller x-coordinate?

  1. x2−10 x+24=(x−4) (x−6)=0\displaystyle x^{2} - 10\, x + 24 = \left(x - 4\right)\, \left(x - 6\right) = 0x2−10x+24=(x−4)(x−6)=0.
  2. The x-values are 4 and 6; the smaller is 4. The corresponding point is (4, 16).

4

WORKED EXAMPLE 2

A rectangle has area xy=40\displaystyle xy = 40xy=40 and x+y=13\displaystyle x + y = 13x+y=13. What is the smaller possible value of x?

  1. x (13−x)=40\displaystyle x\, \left(13 - x\right) = 40x(13−x)=40, so (x−5) (x−8)=0\displaystyle \left(x - 5\right)\, \left(x - 8\right) = 0(x−5)(x−8)=0.
  2. x is 5 or 8; the smaller is 5.

5

Common pitfalls

Possible mix-up: An x-value alone is the full intersection point.

Find its corresponding y-value too.

Possible mix-up: Every line crosses a parabola twice.

A line can touch once or have no real intersection.

Explain it to yourself

In lines & parabolas: nonlinear systems, what information determines your method, and how can you check whether your result fits the original conditions?

Discuss together

Ask your learner to explain one representation and check the conditions in lines & parabolas: nonlinear systems. Compare independent answers with hinted attempts and application problems. A completed lesson is not a PSAT score prediction.

Preview the eight practice prompts
  1. Problem 1

    Find intersections of y=x2\displaystyle y = x^{2}y=x2 and y=6 x−8\displaystyle y = 6\, x - 8y=6x−8. What is the smaller x-coordinate?

    Practise this exact problem →
  2. Problem 2

    The line y=2\displaystyle y = 2y=2 and parabola y=x2+3\displaystyle y = x^{2} + 3y=x2+3 have how many real intersection points?

    Practise this exact problem →
  3. Problem 3

    The curves y=x2\displaystyle y = x^{2}y=x2 and y=6 x−9\displaystyle y = 6\, x - 9y=6x−9 meet. What is the y-coordinate of their intersection?

    Practise this exact problem →
  4. Problem 4

    Two quantities obey xy=10\displaystyle xy = 10xy=10 and x+y=7\displaystyle x + y = 7x+y=7. What is the smaller possible value of x?

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  5. Problem 5New context

    A curved track and straight path follow y=x2\displaystyle y = x^{2}y=x2 and y=8 x−15\displaystyle y = 8\, x - 15y=8x−15. What is the smaller x-coordinate?

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  6. Problem 6New context

    A horizontal platform y=3\displaystyle y = 3y=3 and parabola y=x2+4\displaystyle y = x^{2} + 4y=x2+4 have how many real intersection points?

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  7. Problem 7New context

    A line just touches a curved path y=x2\displaystyle y = x^{2}y=x2 and y=8 x−16\displaystyle y = 8\, x - 16y=8x−16 at one point. What is the y-coordinate of their intersection?

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  8. Problem 8New context

    A rectangle has area xy=18\displaystyle xy = 18xy=18 and x+y=9\displaystyle x + y = 9x+y=9. What is the smaller possible value of x?

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Open stage PSAT 2.5 in the student workspace →

STAGE PSAT 2.6 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Exponential Growth, Decay & Tables

The idea to remember

Equal input intervals give a constant output ratio in an exponential model. A growth factor must match the time unit.

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Goal: Build and interpret exponential models from rates or tables, including decay and different time units.

Before you begin: Read the problem carefully and use the linked foundation lessons when a term or operation is unfamiliar.

PARENT EQUATION

Exponential family: y=bx (b>0,b≠1)\displaystyle y = b^{x}\, \left(b > 0, b \ne 1\right)y=bx(b>0,b=1)

The unshifted graph passes through (0, 1), stays positive and approaches y=0\displaystyle y = 0y=0. It grows for b>1\displaystyle b > 1b>1 and decays for 0<b<1\displaystyle 0 < b < 10<b<1.

How does the question’s equation change this graph?

In y=a⋅bx−h+k\displaystyle y = a \cdot b^{x - h} + ky=a⋅bx−h+k, h shifts horizontally, a scales or reflects vertically, and k moves the horizontal asymptote to y=k\displaystyle y = ky=k.

Check: Keep the base b from the question. Exponential growth multiplies by a factor; it does not add the same amount each step.

First identify this family. Then read the coefficients and shifts in your actual equation; this reference does not replace the question’s given values.

Make sense of the idea

In A (t)=A0 bt\displaystyle A\, \left(t\right) = A_{0}\, b^{t}A(t)=A0​bt, A0\displaystyle A_{0}A0​ is the initial value and b is the factor for one unit of t. Growth of r as a decimal uses b=1+r\displaystyle b = 1 + rb=1+r; decay uses b=1−r\displaystyle b = 1 - rb=1−r.

Choose a method

Equal time steps have equal ratios in an exponential model; a linear model has equal differences. Check that the time steps are equal before comparing.

Check and interpret

If a process doubles every 3 hours and t is hours, use A0⋅2t3\displaystyle A_{0} \cdot 2^{\frac{t}{3}}A0​⋅23t​. The denominator converts elapsed hours into the number of doubling periods. Parent family: y=bx\displaystyle y = b^{x}y=bx, b>0\displaystyle b > 0b>0 and b≠1\displaystyle b \ne 1b=1.

WORKED EXAMPLE 1

A quantity starts at 400 and grows by 20%\displaystyle 20\%20% each year. Find its value after 2 years.

  1. A (2)=400×1.22\displaystyle A\, \left(2\right) = 400 \times 1.2^{2}A(2)=400×1.22.
  2. The value is 576. The increase is compounded, not two increases on the original base.

576

WORKED EXAMPLE 2

A sensor gives values 5, 15, 45 at times 0, 1, 2. Assuming an exponential model, predict its value at time 3.

  1. The ratios are 155=3\displaystyle \frac{15}{5} = 3515​=3 and 4515=3\displaystyle \frac{45}{15} = 31545​=3.
  2. The next value is 45×3=135\displaystyle 45 \times 3 = 13545×3=135; unequal differences distinguish this from a linear model.

135

Common pitfalls

Possible mix-up: A 20%\displaystyle A\, 20\%A20% decrease uses factor 0.20.

It retains 80%\displaystyle 80\%80%, so the factor is 0.80.

Possible mix-up: Doubling every 3 hours means exponent 3t.

The number of periods is t3\displaystyle \frac{t}{3}3t​.

Explain it to yourself

In exponential growth, decay & tables, what information determines your method, and how can you check whether your result fits the original conditions?

Discuss together

Ask your learner to explain one representation and check the conditions in exponential growth, decay & tables. Compare independent answers with hinted attempts and application problems. A completed lesson is not a PSAT score prediction.

Preview the eight practice prompts
  1. Problem 1

    A quantity starts at 200 and grows by 20%\displaystyle 20\%20% each year. Find its value after 2 years.

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  2. Problem 2

    A quantity keeps 80%\displaystyle 80\%80% of its amount each hour. It starts at 300. How much remains after 2 hours?

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  3. Problem 3

    A model doubles every 3 hours, starting at 4. What is its amount after 9 hours?

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  4. Problem 4

    A function gives values 2, 6, 18 at times 0, 1, 2. Assuming an exponential model, predict its value at time 3.

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  5. Problem 5New context

    A habitat population starts at 300 and grows by 20%\displaystyle 20\%20% each year. Find its value after 2 years.

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  6. Problem 6New context

    A filter retains 80%\displaystyle 80\%80% of its amount each hour. It starts at 400. How much remains after 2 hours?

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  7. Problem 7New context

    A culture doubles every 3 hours, starting at 5. What is its amount after 9 hours?

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  8. Problem 8New context

    A sensor gives values 3, 9, 27 at times 0, 1, 2. Assuming an exponential model, predict its value at time 3.

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