Mixed practice: Advanced Math

B: 3x² − 10x − 8

Distribute both terms: (3x + 2)(x − 4) = 3x² − 12x + 2x − 8 = 3x² − 10x − 8.

The middle coefficient comes from two products, not one. A fast check at x = 0 gives −8, eliminating choices with the wrong constant, but it does not distinguish all remaining choices. Check at a second convenient input or complete the expansion. Review 1.02.

C: 12 The expression is already in vertex form. Since (x − 2)² ≥ 0, multiplying by −3 makes that term nonpositive. Therefore −3(x − 2)² + 12 ≤ 12, with equality at x = 2. The maximum output is 12. The numbers 2 and −3 are respectively the vertex input and vertical-scale coefficient; neither is the requested value. Review 3.06.

256

Substitute t = 6 into the whole exponent:

Q(6) = 500(0.8)^6/2 = 500(0.8)³ = 500(0.512) = 256.

There are three 2-unit decay periods, each retaining 80% of the preceding amount. Applying the factor six times would use the wrong time interval. Estimation also helps: the result should be positive and below 500, but well above 500(0.8)⁶. Review 5.03.

D: $\sqrt{\text{a}}$

The positive-base assumption makes rational exponent rules straightforward. Divide powers with the same base by subtracting exponents:

$\frac{{\text{a}}^{\text{5/6}}}{{\text{a}}^{\text{1/3}}}$ = a^5/6−2/6 = a^3/6 = a^1/2 = $\sqrt{\text{a}}$.

Do not divide the exponents; their difference is required by the quotient rule. Also, the rule is not an instruction to subtract the actual numerator and denominator. Review 2.09.

12

Both sides of $\sqrt{\text{4x + 1}}$ = 7 are nonnegative. Square, solve, and verify:

4x + 1 = 49, 4x = 48, x = 12.

The original radicand is 49, so $\sqrt{\text{49}}$ = 7 as required. Squaring creates candidate solutions; substitution confirms this one. There is no extra negative branch for the value of the principal square root itself. Review 4.07.

A: (−2, 2)

The given point means f(2) = −3. To use that known output in g(x) = f(x + 4) + 5, make the inside input equal 2: x + 4 = 2 x = −2.

Then g(−2) = f(2) + 5 = −3 + 5 = 2. Thus the new point is (−2, 2). Solving the inside equation avoids the common sign error of shifting right when the formula contains x + 4. Review 6.05.

16

At an intersection, x² = 3x + 4. Rearrange and factor:

x² − 3x − 4 = (x − 4)(x + 1) = 0.

The inputs are 4 and −1. The question selects the positive input and asks for its y-coordinate. Thus y = 4² = 16. Check the other equation: 3(4) + 4 = 16. Entering 4 would solve the intermediate equation but answer the wrong question. Review 7.01.

B: 3(2x − 3)(2x + 3)

Remove the greatest common factor, then factor the difference of squares:

12x² − 27 = 3(4x² − 9) = 3((2x)² − 3²) = 3(2x − 3)(2x + 3).

A repeated factor would produce a nonzero linear term. Multiplying conjugates cancels the cross terms, matching the absence of an x term in the original polynomial. Review 1.03 and 1.04.

D: f(x) = 2(x + 1)(x − 5)

The zeros determine factors x + 1 and x − 5, but not the leading coefficient. Write f(x) = a(x + 1)(x − 5) and use the given output: −10 = f(0) = a(1)(−5) = −5a, a = 2.

The resulting function has both required zeros and the correct y-intercept. An answer with the right roots but the wrong scale is not enough. Review 3.08.

−2

Record x ≠ 2 before factoring. The numerator is (x−2)(x+2). A rational expression is zero when its numerator is zero and its denominator is nonzero. The numerator candidates are 2 and −2, but 2 is excluded. At −2, the denominator is −4 and the numerator is zero, so the quotient is zero. Equivalently, simplify to x + 2 while retaining x ≠ 2. Review 4.02.

9

A negative exponent takes a reciprocal:

($\frac{\text{1}}{\text{27}}$)^−2/3 = 27^2/3 = ($\sqrt[\text{3}]{\text{27}}$)² = 3² = 9.

The negative sign in the exponent does not turn the output negative. An alternative common-base route is (3⁻³)^−2/3 = 3². Both methods are valid because the base is positive. Review 2.03.

C: 224

The problem explicitly identifies f as exponential. Consecutive outputs double, so b = 2, while f(0) = A = 7. Therefore f(t) = 7 ⋅ 2^t and f(5) = 7 ⋅ 2⁵ = 7(32) = 224.

Or extend the table: f(4) = 112 and f(5) = 224. The input 0 is the initial row, not the first multiplication period. A finite table alone would not uniquely specify an arbitrary function; the stated model matters. Review 5.05.

A: 5

Real square-root outputs require x − 5 ≥ 0, so x ≥ 5. The endpoint x = 5 is allowed and gives f(5) = 2. Thus the smallest input in the domain is 5. The outside +2 changes output values, not the square root’s input restriction. Confusing the minimum output 2 with the smallest allowed input 5 produces a plausible distractor. Review 6.09.

D: (2, 3)

Testing choices is efficient because the target is a single offered pair. For (2, 3),

2² + 3² = 4 + 9 = 13, 3 = 2 + 1.

Both equations hold. The other listed pairs all satisfy the circle equation but fail the line equation. A system requires simultaneous truth, so checking only the easier equation is insufficient. Algebraically, substitution gives 2x² + 2x − 12 = 0, with intersections (2, 3) and (−3, −2); only the first is listed. Review 7.06.

3

Absolute value equal to a positive constant gives two branches:

3x + 2 = 11 or 3x + 2 = −11.

These yield x = 3 and x = −13/3. The greater solution is 3. Both are valid because their inside expressions have magnitude 11. The question asks for one selected solution, so entering both would not follow the target. Review 4.10.

B: 2

Expand the left side: (2x + a)(x − 3) = 2x² + (a − 6)x − 3a.

The constant identity gives −3a = −12, so a = 4. The linear coefficient is then b = a − 6 = −2. Therefore a + b = 4 − 2 = 2. Comparing only the leading coefficient provides no new information because it is already 2 on both sides. Review 1.11.

5

The two given inputs are three units apart. Dividing the model values cancels the initial coefficient:

$\frac{\text{f(4)}}{\text{f(1)}}$ = b³ = $\frac{\text{80}}{\text{10}}$ = 8.

Because b > 0, b = 2. Then 10 = Ab gives A = 5. Finally, f(0) = Ab⁰ = A = 5. A factor of 8 belongs to the whole three-unit interval, not to a single unit. Review 5.06.

C: −7

Complete the square, keeping the leading coefficient outside the adjusted square:

3x² − 18x + 20 = 3(x² − 6x) + 20 = 3(x − 3)² − 27 + 20 = 3(x − 3)² − 7.

The squared term is nonnegative, so the minimum output is −7, achieved at x = 3. The vertex formula also gives x = −(−18)/(2 ⋅ 3) = 3, after which substitution gives −7. Review 3.07.

A: 5

Use the common positive base 5:

25^x+1 = 5^2x+2, 125^x−1 = 5^3x−3.

Since powers of base 5 are one-to-one, 2x + 2 = 3x − 3, so x = 5. Check the exponents: both sides equal 5¹². Equating x + 1 and x − 1 directly would ignore the different original bases. Review 2.10 and 5.04.

D: (12, −11)

Use the known value f(6) = 4. The inside input in g equals 6 when x/2 = 6, so the new input is 12. The new output is −3(4) + 1 = −11. Therefore g(12) = −11. The factor 1/2 inside the function stretches horizontal distances by 2; it does not halve the point’s input. Review 6.07.

−9

Set the equations equal and collect terms:

x² = 6x + k x² − 6x − k = 0.

The leading coefficient is always nonzero, so one intersection requires D = 36 + 4k = 0, giving k = −9. The equation becomes (x − 3)² = 0, and the point is (3, 9). The number of distinct roots, not their multiplicity, determines the number of intersections here. Review 7.12.

B: the amount above 40 decreases by 10% per unit

Subtract the baseline: Q(t) − 40 = 100(0.9)^t.

Therefore Q(t + 1) − 40 = 0.9[Q(t) − 40]. The excess above 40, not the entire amount, is multiplied by 0.9. For example, Q(0) = 140 and Q(1) = 130; that is not a 10% decrease of 140. The excess remains positive for finite t, so the model does not reach exactly 40 at t = 10. Review 5.14.

−12

The factor condition gives P(2) = 0: 8 + 2a − 10 = 0, a = 1.

Division by x + 1 = x − (−1) leaves remainder P(−1), so

P(−1) = (−1)³ + (−1) − 10 = −12.

Use −1, not 1, in the remainder evaluation. The factor condition and the requested divisor involve different inputs and different purposes. Review 1.12.

C: no solution

The original equation requires x ≠ 3. On this allowed domain, multiplication by x − 3 gives 2 = x − 1, hence the only candidate is x = 3. It is excluded, so there is no solution. Neither original fraction has a value there. This is a contradiction on the original domain, even though the cleared linear equation by itself has a solution. Review 4.05.

B: $\frac{\text{13}}{\text{2}}$

From the quadratic coefficients,

r + s = $\frac{\text{12}}{\text{4}}$ = 3, rs = $\frac{\text{5}}{\text{4}}$.

Then r² + s² = (r + s)² − 2rs = 9 − 5/2 = 13/2. The product enters with a factor of 2 because expanding (r + s)² creates two cross terms. Finding the individual roots is possible but unnecessary. Review 3.13.

A: [−7, 2] The vertex of 2 − (x − 1)² is (1, 2), and its input lies in [−1, 4], so the maximum is 2. At the endpoints,

f(−1) = 2 − 4 = −2, f(4) = 2 − 9 = −7.

The minimum is −7. The continuous quadratic takes every value between these extremes, so the range is [−7, 2]. The unrestricted range (−∞, 2] ignores the stated domain. Review 6.10.

D: 64

Rewrite the target as a power of the given quantity:

27^2t = 3^6t = (3^2t)³ = (9^t)³ = 4³ = 64.

There is no need to determine t. The key is matching the target exponent to a multiple of the known exponent. Treating the ratio of the bases, 27/9 = 3, as an output multiplier would not respect exponentiation. Review 2.12.

C: 21

Square the sum and use the given sum of squares:

(x + y)² = x² + y² + 2xy, 100 = 58 + 2xy.

Thus 2xy = 42 and xy = 21. Solving separately for x and y would add work without changing the target. As a check, the pair 3, 7 has sum 10, squared sum 58, and product 21. Review 7.15 and 1.13.

3

When k = 0, the equation is linear: −6x + 3 = 0, with one solution. For k ≠ 0, one distinct real root requires

D = (−6)² − 4(k)(3) = 36 − 12k = 0, k = 3.

At k = 3, the equation is 3(x − 1)² = 0. The qualifying parameters are 0 and 3, whose sum is 3. Check the degree before using a quadratic-only criterion. Review 3.15.

A: 3z²$\sqrt{\text{2}}$

Extract the perfect square: $\sqrt{{\text{18z}}^{\text{4}}}$ = 3$\sqrt{\text{2}}$$\sqrt{\text{(}{\text{z}}^{\text{2}}{\text{)}}^{\text{2}}}$ = 3$\sqrt{\text{2}}$ |z²| = 3z²$\sqrt{\text{2}}$.

Because z² ≥ 0 for every real z, |z²| = z². Absolute value is essential in the general rule $\sqrt{{\text{u}}^{\text{2}}}$ = |u|, but here the quantity being squared is already nonnegative. The other options have the wrong coefficient or power. Review 2.07.

B: 7^5/3 For a pure exponential, a change of h input units multiplies the output by b^h. The condition f(t + 3) = 7f(t) gives b³ = 7, so b = 7^1/3 since b > 0. Therefore

$\frac{\text{f(t + 5)}}{\text{f(t)}}$ = b⁵ = 7^5/3.

The exponent is the desired five-unit interval divided by the known three-unit interval. A linear proportional adjustment of the multiplier would be incorrect. Review 5.12.

7

The nonnegative square root requires x − 2 ≥ 0, so x ≥ 2. Squaring gives

3x + 4 = (x − 2)² = x² − 4x + 4, x(x − 7) = 0.

The candidates are 0 and 7. Reject 0 because it would make the right side negative. At 7, $\sqrt{\text{25}}$ = 5 = 7 − 2. Squaring alone loses sign information, which the original restriction restores. Review 4.08.

C: (x − 2)(x + 2)(x − 3)(x + 3) Temporarily treat x² as one object. The quadratic pattern is

x⁴ − 13x² + 36 = (x² − 4)(x² − 9).

Each factor is a difference of squares, giving (x−2)(x+2)(x−3)(x+3). Check the middle term before finishing: −4x²−9x² = −13x². The square roots of 4 and 9, not the numbers 4 and 9 themselves, produce the linear factors. Review 1.15.

A: (−1, −1)

Read the original vertex as (−2, 3). In g(x) = f(x − 1) − 4, the inside change shifts the graph right 1, and the outside change shifts it down 4. Thus the vertex becomes

(−2 + 1, 3 − 4) = (−1, −1).

No scale change or reflection occurs, so a vertex maps to a vertex. The labeled point (0, 11) is consistent with the graph but is not needed for this target. Review 6.05.

1

Equating the outputs and collecting terms gives

x² + 3x = (k + 1)x² − x + 4 kx² − 4x + 4 = 0.

At k = 0, this is a linear equation with one solution. At k ≠ 0, one solution requires 16 − 16k = 0, so k = 1. Each valid x determines exactly one y in the original functions. The two parameter values sum to 1. Notice that the difference can become linear even when both original rules remain quadratic. Review 7.13.

5

Use a difference of squares, or expand just enough to compare coefficients:

(x + a)² − (x + b)² = 2(a − b)x + (a² − b²).

Hence a − b = 4 and (a − b)(a + b) = 24, so a + b = 6. Solving gives a = 5 and b = 1, therefore ab = 5. An equally direct target identity is ab = [(a + b)² − (a − b)²]/4 = (36 − 16)/4 = 5. Review 1.11 and 1.13.

D: 10%

The base 0.81 is the multiplier for a two-year interval. Rewrite the model with exponent t:

300(0.81)^t/2 = 300 ($\sqrt{\text{0.81}}$)^t = 300(0.9)^t.

The amount retains 90% each year, so it decreases by 10% per year. A 19% decrease applies over two years; dividing 19 by 2 would not account for compounding. Review 5.07.

A: 48

The roots 2 and 8 are symmetric around the vertex input 5. Evaluate the factored rule at that midpoint:

f(5) = a(5 − 2)(5 − 8) = −9a = −27, a = 3.

Then f(0) = 3(−2)(−8) = 48. Because a > 0, the vertex output is indeed the minimum. The minimum itself is not the requested y-intercept. Review 3.08.

4

Recognize the product exponent as a power of a power. The bases are positive, so the real-exponent rule applies:

2^xy = (2^x)^y = 3^y = 4.

The supplied relationships already form a substitution chain. Solving for either exponent separately is unnecessary. This is a useful reminder that a problem containing unknown exponents may ask for a quantity determined without calculating those exponents. Review 2.12 and 2.15.

D: a hole at (3, $\frac{\text{4}}{\text{3}}$)

Factor before classifying excluded inputs:

f(x) = $\frac{\text{(x + 5)(x − 3)}}{\text{(x − 3)(x + 3)}}$ = $\frac{\text{x + 5}}{\text{x + 3}}$, x ≠ 3, −3.

The canceled factor creates a missing point at x = 3. The reduced rule gives its would-be output (3+5)/(3+3) = 4/3. The uncanceled denominator factor produces a vertical asymptote at x = −3, not at 3. Neither excluded input is an x-intercept. Review 4.02.

B: 7

Equal outputs at x = −1 and x = 5 locate the quadratic’s axis halfway between them: h = 2. Since the minimum is −1, write f(x) = a(x − 2)² − 1. Use f(−1) = 17:

17 = 9a − 1, a = 2.

Therefore f(0) = 2(0 − 2)² − 1 = 7. The symmetry conclusion uses the stated quadratic model with a nonzero leading coefficient; equal outputs in an arbitrary function would not establish this form. Review 6.15.

C: 10

At an intersection, x² − 4x + 10 = 2x + c x² − 6x + 10 − c = 0.

Exactly one point requires D = 36 − 4(10 − c) = 4c − 4 = 0, giving c = 1. The repeated root is x = 6/2 = 3, and y = 2(3) + 1 = 7. Hence x + y = 10. The parameter, input, output, and requested sum are four different quantities; keep them labeled. Review 7.03 and 7.12.