FL Algebra 1 EOC Practice Questions (2026) — Free Sample Questions with Answers

Answer: A

1. 1. $f$ has vertex $(0,-4)$.
2. 2. $g(x)=f(x+3)$ replaces $x$ with $x+3$: every point moves 3 units left.
3. 3. The vertex of $g$ is $(-3,-4)$; check: $g(-5)=f(-2)=0$ and $g(-1)=f(2)=0$.

Graph A: Correct. Graph A has its vertex at $(-3,-4)$ and x-intercepts $-5$ and $-1$: the graph of $f$ shifted 3 units left, since $f(x+3)=(x+3{)}^2-4$.

Graph B: Graph B has its vertex at $(3,-4)$, a shift 3 units right; that is $f(x-3)$. Adding 3 inside the function moves the graph left.

Graph C: Graph C has its vertex at $(0,-1)$, a shift up 3; that is $f(x)+3$, with the 3 outside the function.

Graph D: Graph D keeps the vertex at $(0,-4)$ but is much narrower; that is $f(3x)=9x^2-4$, multiplying the input by 3 instead of adding 3.

$f(x+k)$ shifts the graph horizontally by $-k$ (left when $k>0$); $f(x)+k$ shifts vertically; $f(kx)$ compresses or stretches horizontally.

Answer: A

1. 1. y-intercepts: $f(0)=20$; $g(0)=2^0=1$.
2. 2. $f$ has slope $\frac{26-20}{2}=3$, so $f(x)=3x+20$.
3. 3. Compare: $f(5)=35>g(5)=32$, but $f(6)=38<g(6)=64$. After that, $g$ doubles each unit while $f$ grows by 3.

The y-intercept of $f$ is greater, and $g(x)$ eventually exceeds $f(x)$.: Correct. $f(0)=20>g(0)=1$, but $g$ doubles while $f$ adds 3 per unit: $g(6)=64>f(6)=38$, and $g$ stays greater for all larger $x$.

The y-intercept of $f$ is greater, and $f(x)$ stays greater than $g(x)$ for all $x>0$.: True at first ($f(4)=32>16$), but by $x=6$, $g(6)=64$ is already greater than $f(6)=38$. A growing exponential always overtakes a linear function.

The y-intercept of $g$ is greater, and $g(x)$ eventually exceeds $f(x)$.: The y-intercept of $g$ is $2^0=1$, not 2; $f$'s y-intercept, 20, is greater.

The y-intercept of $g$ is greater, and $f(x)$ stays greater than $g(x)$ for all $x>0$.: Both parts are false: $g(0)=1<20$, and $g$ overtakes $f$ by $x=6$.

An increasing exponential function eventually exceeds any linear (or quadratic) function, regardless of the starting values.

Answer: B

1. 1. In $V=84.8(1.06{)}^{t-1}$, the coefficient 84.8 is the output when the exponent $t-1$ equals 0, that is, when $t=1$.
2. 2. Compare with Hiroshi's model at $t=1$: $80(1.06{)}^1=84.8$. ✓
3. 3. The two models agree for every $t$: $84.8(1.06{)}^{t-1}=80(1.06)(1.06{)}^{t-1}=80(1.06{)}^t$.
4. 4. So 84.8 is the value, in dollars, 1 year after Hiroshi buys the card; the value at purchase is still 80.

the amount, in dollars, that the value increases each year: 84.8 is a value, not an increase. The increase in the first year is $84.8-80=4.8$ dollars, and the yearly increase grows each year because the growth is exponential.

the value, in dollars, of the card 1 year after Hiroshi buys it: Correct. When $t=1$, the exponent $t-1$ is 0, so $V=84.8(1.06{)}^0=84.8$, the value 1 year after purchase. This matches Hiroshi's model: $80(1.06{)}^1=84.8$.

the value, in dollars, of the card when Hiroshi buys it: The value at purchase ($t=0$) is 80 dollars. In the sister's model, $t=0$ gives $84.8(1.06{)}^{-1}=80$, so 84.8 is not the starting value.

the value, in dollars, of the card 1 year before Hiroshi buys it: Reads $t-1$ as "one year earlier." 84.8 is the output when $t-1=0$, that is, at $t=1$; one year before purchase ($t=-1$) the model gives about 75.47 dollars.

In $a(1+r{)}^{t-c}$, the coefficient $a$ is the value when $t=c$ (where the exponent is 0), which is not always the starting value.

Answer: D

1. 1. ${\left(\frac{1}{8}\right)}^{-\frac{2}{3}}=8^{\frac{2}{3}}={\left(\sqrt[3]{8}\right)}^2=2^2=4$.
2. 2. $16^{\frac{3}{4}}={\left(\sqrt[4]{16}\right)}^3=2^3=8$.
3. 3. $4\cdot 8=32=2^5$, so Step 4 should read $2^5$, not $2^6$.

Step 1: Step 1 is correct: a negative exponent on $\frac{1}{8}$ means taking the reciprocal, so ${\left(\frac{1}{8}\right)}^{-\frac{2}{3}}=8^{\frac{2}{3}}$.

Step 2: Step 2 is correct: $a^{\frac{m}{n}}={\left(\sqrt[n]{a}\right)}^m$, so the denominators 3 and 4 become the root indexes.

Step 3: Step 3 is correct: $\sqrt[3]{8}=2$ and $\sqrt[4]{16}=2$.

Step 4: Correct. When multiplying powers with the same base, the exponents are added: $2^2\cdot 2^3=2^5=32$. Amara multiplied the exponents to get $2^6$.

Product of powers with the same base: add the exponents ($a^m\cdot a^n=a^{m+n}$); multiplying exponents is the rule for a power of a power.

Answer: B

1. Tank M rate: $\frac{335-395}{10-5}=\frac{-60}{5}=-12$ gallons per minute.
2. Tank M at $t=0$: $395+5(12)=455$ gallons.
3. Tank N: starts at 450 gallons and changes by $-15$ gallons per minute.
4. So Tank M started with more (455 > 450), and Tank N drains faster (15 > 12).

Tank M started with more water, and Tank M drains faster.: Tank M loses 60 gallons per row, but each row is 5 minutes, so its rate is 12 gallons per minute, slower than Tank N's 15.

Tank M started with more water, and Tank N drains faster.: Correct. Tank M started with $395+60=455$ gallons, more than 450, and drains 12 gallons per minute, slower than Tank N's 15.

Tank N started with more water, and Tank M drains faster.: Reads 395 (the value at 5 minutes) as Tank M's starting amount and uses the change per row, 60, as its rate.

Tank N started with more water, and Tank N drains faster.: Tank N does drain faster, but Tank M's starting amount is 455 gallons (work back 5 minutes), not 395.

Find the rate of change as change in output over change in input, and the initial value by working back to t = 0, before comparing representations.

Answer: D

1. $2x+3y=12$: intercepts $(6,0)$ and $(0,4)$; solid. Test $(0,0)$: $0\le 12$ is true, so shade the origin side.
2. $x-y>1$ becomes $y<x-1$: dashed line through $(0,-1)$ and $(1,0)$; shade below.
3. The solution set is the overlap below both lines.
4. Check with $(3,0)$: $6\le 12$ and $3>1$ are both true, and $(3,0)$ is shaded in Graph D.

Graph A: Shades above $y=x-1$. Solving $x-y>1$ gives $-y>-x+1$, and dividing by $-1$ reverses the symbol: $y<x-1$, the region below.

Graph B: The solid line crosses at $(4,0)$ and $(0,6)$, which swaps the intercepts of $2x+3y=12$; the correct intercepts are $(6,0)$ and $(0,4)$.

Graph C: The line styles are switched: $2x+3y\le 12$ needs a solid line and $x-y>1$ a dashed one.

Graph D: Correct. Solid line through $(6,0)$ and $(0,4)$ shaded toward the origin, and dashed line $y=x-1$ shaded below; the overlap is shaded.

Graph each inequality from intercepts or slope-intercept form, reverse the symbol when dividing by a negative, and keep only the overlapping region.

Answer: A

1. x-intercept: $3x=6$, so $(2,0)$. y-intercept: $-2y=6$, so $(0,-3)$.
2. The symbol is $\ge$, so the boundary is solid.
3. Test $(0,0)$: $3(0)-2(0)=0\ge 6$ is false, so shade the side that does not contain the origin.
4. Equivalent slope-intercept form: $y\le \frac{3}{2}x-3$ (shade below), which matches Graph A.

Graph A: Correct. Solid line through the intercepts $(2,0)$ and $(0,-3)$, shaded on the side away from the origin, because $(0,0)$ gives $0\ge 6$, which is false.

Graph B: Right boundary, but the shading includes the origin. Solving for $y$ means dividing by $-2$, which reverses the symbol to $y\le \frac{3}{2}x-3$; keeping $\ge$ shades the wrong side.

Graph C: Swapped intercepts: this line crosses at $(-3,0)$ and $(0,2)$. Setting $y=0$ in $3x-2y=6$ gives $x=2$, and setting $x=0$ gives $y=-3$.

Graph D: Right boundary and side, but dashed; $\ge$ includes the boundary, so the line must be solid.

Graph Ax + By ≥ C from its intercepts, then pick the side with a test point; if you solve for y, reverse the symbol when dividing by a negative coefficient.

Answer: B

1. 1. Subtract 3 from all three parts: $-10<-2x\le 6$.
2. 2. Divide all three parts by $-2$ and reverse both signs: $5>x\ge -3$.
3. 3. Rewrite: $-3\le x<5$ — closed at $-3$ (from $\le$), open at 5 (from $<$).
4. 4. Test $x=0$: $3-0=3$ and $-7<3\le 9$ is true.

Graph A: Graph A shows $-5<x\le 3$: it divides by 2 instead of $-2$ and never reverses the signs, so both endpoints have the wrong signs.

Graph B: Correct. Graph B shows a closed dot at $-3$, an open dot at 5 and the segment between: $-3\le x<5$.

Graph C: Graph C shows $-3<x\le 5$: the endpoints are right but the dots are switched. The $\le 9$ part gives $x\ge -3$ (closed dot), and the $-7<$ part gives $x<5$ (open dot).

Graph D: Graph D shows rays $x\le -3$ or $x>5$; reversing the inequality signs does not turn an "and" inequality into an "or" inequality.

Dividing a compound inequality by a negative reverses both signs; each endpoint keeps the inclusion (open or closed) of the inequality it came from.

Answer: B

1. 1. x-intercept: $y=0$ gives $2x=9$, $x=4.5$.
2. 2. y-intercept: $x=0$ gives $-3y=9$, $y=-3$.
3. 3. Slope check: $y=\frac{2}{3}x-3$ rises 2 for every 3 to the right.

Graph A: Graph A has intercepts $(4.5,0)$ and $(0,3)$; it treats $-3y=9$ as $y=3$, dropping the negative sign.

Graph B: Correct. Graph B crosses the x-axis at $(4.5,0)$ and the y-axis at $(0,-3)$: $2x=9$ gives $x=4.5$ and $-3y=9$ gives $y=-3$.

Graph C: Graph C has intercepts $(-3,0)$ and $(0,4.5)$; it swaps the x- and y-intercepts.

Graph D: Graph D has intercepts $(-4.5,0)$ and $(0,-3)$; the sign of the x-intercept is wrong.

For $Ax+By=C$, set $y=0$ for the x-intercept and $x=0$ for the y-intercept — keep the sign of each coefficient.

Answer: (1) 2.5, (2) −3/4, (3) −4

1. 1. Slope: $\frac{-3.5-2.5}{4-(-4)}=-\frac{6}{8}=-\frac{3}{4}$.
2. 2. Point-slope with $(-4,2.5)$: $y-2.5=-\frac{3}{4}(x-(-4))$.
3. 3. Check the other point: $-3.5-2.5=-6$ and $-\frac{3}{4}(4+4)=-6$.

In $y-y_1=m(x-x_1)$ the coordinates are subtracted, so a negative coordinate appears as $x-(-4)$, which is $x+4$.

Answer: D

1. $f(0)=8$: the y-intercept is 8.
2. $f(1)=4$, $f(2)=2$, $f(3)=1$: the outputs halve each step.
3. The base $\frac{1}{2}$ is less than 1, so the graph decreases and approaches $y=0$.

Graph A: Passes through $(0,8)$ but increases; a base of $\frac{1}{2}$ makes the function decrease.

Graph B: A line through $(0,8)$ and $(2,0)$: it subtracts 4 each unit instead of halving, and it crosses the x-axis, which an exponential function never does.

Graph C: Swaps $a$ and $b$: this is $\frac{1}{2}(8{)}^x$, with y-intercept $\frac{1}{2}$ and rapid growth.

Graph D: Correct. y-intercept 8, passing through $(1,4)$, $(2,2)$ and $(3,1)$, decreasing and approaching the asymptote $y=0$.

For ab^x, the y-intercept is a, each unit step multiplies by b, and the graph approaches the horizontal asymptote y = 0 without crossing it.

Answer: D

1. Solve $-4.9t^2+14.7t+1.2=0$: $a=-4.9$, $b=14.7$, $c=1.2$.
2. $b^2-4ac=216.09+23.52=239.61$, and $\sqrt{239.61}\approx 15.479$.
3. $t=\frac{-14.7\pm 15.479}{-9.8}$: $t\approx -0.08$ or $t\approx 3.08$.
4. Time must be positive, so $t\approx 3.08$ seconds. Check: $h(3.08)\approx -46.48+45.28+1.2\approx 0$. ✓

$-0.08$: This root of the equation is negative; time after the hit cannot be negative, so it is not viable.

$1.50$: This is the time of the maximum height, $t=-\frac{14.7}{2(-4.9)}$, not the time the height is 0.

$3.00$: Ignores the starting height 1.2 m: $-4.9t^2+14.7t=0$ gives $t=3$, but the ball starts above the ground.

$3.08$: Correct. Solving $h(t)=0$ with the quadratic formula gives $t\approx 3.08$ or $t\approx -0.08$; only the positive time is viable.

To find when an object lands, solve h(t) = 0 and keep only the solution that fits the context (here, positive time).

Answer: B

1. Vertex form $a(x-h{)}^2+k$ with $h=9$, $k=320$: the vertex is $(9,320)$.
2. $a=-5<0$, so the parabola opens downward and the vertex is the maximum.
3. Meaning: charging $9 per meal gives the greatest daily profit, $320.

The daily profit is $9 when the meal price is $320.: Swaps the coordinates: in $(9,320)$, 9 is the price ($x$) and 320 is the profit ($P$).

The greatest daily profit, $320, occurs when the meal price is $9.: Correct. The vertex is $(9,320)$, and because $a=-5<0$ it is a maximum: the highest profit, $320, at a price of $9.

The daily profit is $320 when the meal price is $0.: Treats the vertex as the y-intercept; $P(0)=-5(81)+320=-85$.

The least daily profit, $320, occurs when the meal price is $9.: The parabola opens downward ($a<0$), so the vertex is a maximum, not a minimum.

In vertex form a(x − h)² + k, the vertex (h, k) is a maximum when a < 0 and a minimum when a > 0; interpret h and k with their units.

Answer: D

1. Add 5: $2(x-3{)}^2=50$.
2. Divide by 2: $(x-3{)}^2=25$.
3. Square root: $x-3=5$ or $x-3=-5$, so $x=8$ or $x=-2$.
4. Check $x=-2$: $2(25)-5=45$. ✓

$\{-22,28\}$: Skips the square root: $x-3=\pm 25$ uses $(x-3{)}^2=25$ as if it were $x-3=25$.

$\{-8,2\}$: Sign error: $x-3=\pm 5$ gives $x=3\pm 5$, not $-3\pm 5$.

$\left\{3-5\sqrt{2},3+5\sqrt{2}\right\}$: Does not divide by 2: $(x-3{)}^2=50$ comes from $2(x-3{)}^2=50$ without dividing.

$\{-2,8\}$: Correct. $2(x-3{)}^2=50$, so $(x-3{)}^2=25$, $x-3=\pm 5$, and $x=8$ or $x=-2$.

Isolate the squared expression (undo addition, then multiplication), take both square roots, then solve the linear equations.

Answer: D

1. Group: $(x^3-5x^2)+(-4x+20)=x^2(x-5)-4(x-5)$.
2. Factor out $(x-5)$: $(x-5)(x^2-4)$.
3. Difference of squares: $(x-5)(x-2)(x+2)$.
4. Check with $x=1$: $1-5-4+20=12$ and $(-4)(-1)(3)=12$. ✓

$(x-5)(x^2+4)$: Sign error in the second group: $-4x+20=-4(x-5)$, which gives $x^2-4$, not $x^2+4$.

$(x-5)(x^2-4)$: Equivalent but not complete: $x^2-4$ is a difference of squares and still factors as $(x-2)(x+2)$.

$(x-5)(x-2{)}^2$: Treats $x^2-4$ as a perfect square; it is a difference of squares, $(x-2)(x+2)$.

$(x-5)(x-2)(x+2)$: Correct. Grouping gives $(x-5)(x^2-4)$, and $x^2-4=(x-2)(x+2)$.

Group terms in pairs, factor each pair so the same binomial appears, then factor any remaining difference of squares.

Answer: C

1. $2x\cdot (x^2+4x-1)=2x^3+8x^2-2x$.
2. $-3\cdot (x^2+4x-1)=-3x^2-12x+3$.
3. Add: $2x^3+(8x^2-3x^2)+(-2x-12x)+3=2x^3+5x^2-14x+3$.
4. Check with $x=1$: $(-1)(4)=-4$ and $2+5-14+3=-4$. ✓

$2x^3+5x^2-14x-3$: Sign slip on the last product: $(-3)(-1)=+3$, not $-3$.

$2x^3+11x^2+10x-3$: Multiplies by $+3$ instead of $-3$, which changes the signs of the second row of products.

$2x^3+5x^2-14x+3$: Correct. $2x(x^2+4x-1)=2x^3+8x^2-2x$ and $-3(x^2+4x-1)=-3x^2-12x+3$; combining gives $2x^3+5x^2-14x+3$.

$2x^3+5x^2+10x+3$: Combines $-2x$ and $-12x$ as $+10x$; both terms are negative, so they total $-14x$.

Multiply every term of the first polynomial by every term of the second (here 6 products), then combine like terms, watching each sign.