The most difficult of the Next-Generation Accuplacer math test covers Advanced Algebra & Functions. Try our free Accuplacer Advanced Algebra practice test to review these topics. This test includes 20 multiple choice questions.

Congratulations - you have completed .

You scored %%SCORE%% out of %%TOTAL%%.

Your performance has been rated as %%RATING%%

Your answers are highlighted below.

Question 1 |

### The length of a rectangle is four more than its width. If the area of the rectangle is 12, find the length of the rectangle.

$-6$ | |

$2$ | |

$4.5$ | |

$6$ |

Question 1 Explanation:

The correct choice is (D). Since the length of the rectangle is four more than the rectangle’s width, we can write $l$ as $w + 4$.

Substituting in $w + 4$ for $l$ and $12$ for $A$ into the equation $A = l × w$, we get $12 = w(w + 4)$.

Simplifying, we get $12 = w^2 + 4w$.

Subtracting $12$ from both sides, we get $0 = w^2 + 4w - 12$, which can be factored as $(w + 6)(w - 2) = 0$.

Since $w$ must be positive, we equate $w - 2$ to $0$ to get $w = 2$. Since $l = w + 4$, we find that $l = 6$.

Substituting in $w + 4$ for $l$ and $12$ for $A$ into the equation $A = l × w$, we get $12 = w(w + 4)$.

Simplifying, we get $12 = w^2 + 4w$.

Subtracting $12$ from both sides, we get $0 = w^2 + 4w - 12$, which can be factored as $(w + 6)(w - 2) = 0$.

Since $w$ must be positive, we equate $w - 2$ to $0$ to get $w = 2$. Since $l = w + 4$, we find that $l = 6$.

Question 2 |

### The graph above shows a cost comparison for two different taxi services. The $x$-axis of the graph represents the number of kilometers driven, and the $y$-axis represents the taxi fare in dollars. The red line pictured above represents Taylor’s taxi service and the blue line represents Chandan’s carpool service. How much more does Chandan’s carpool service cost than Taylor’s taxi service per kilometer?

$\$1.75$ | |

$\$2.00$ | |

$\$3.50$ | |

$\$7.00$ |

Question 2 Explanation:

The correct choice is (A). We first find the equation of Chandan’s taxi service. First we find the slope, which is equal to:

$\dfrac{7 - 0 \, \text{ dollars}}{2 - 0 \, \text{ kilometers}}$ $= 3.5 \dfrac{\$}{\text{km}}$ $\text{or } \$3.5 \text{ per km}$

Next, we find the equation of Taylor’s taxi service. We find the slope, which is equal to:

$\dfrac{7 - 0 \, \text{ dollars}}{4 - 0 \, \text{ kilometers}}$ $= 1.75 \dfrac{\$}{\text{km}}$ $\text{or } \$1.75 \text{ per km}$

To find how much more Chandan’s taxi service costs than Taylor’s taxi service, we find $3.5 - 1.75 = \$1.75$ per kilometer.

$\dfrac{7 - 0 \, \text{ dollars}}{2 - 0 \, \text{ kilometers}}$ $= 3.5 \dfrac{\$}{\text{km}}$ $\text{or } \$3.5 \text{ per km}$

Next, we find the equation of Taylor’s taxi service. We find the slope, which is equal to:

$\dfrac{7 - 0 \, \text{ dollars}}{4 - 0 \, \text{ kilometers}}$ $= 1.75 \dfrac{\$}{\text{km}}$ $\text{or } \$1.75 \text{ per km}$

To find how much more Chandan’s taxi service costs than Taylor’s taxi service, we find $3.5 - 1.75 = \$1.75$ per kilometer.

Question 3 |

### Consider the polynomial $2x^7 + 5x^3 + 4x + 3$. Find the value of $3d - c + p$, where $d$ is the degree of the polynomial, $c$ is the constant term of the polynomial, and $p$ is the product of the coefficients of the non-constant terms of the polynomial.

$147$ | |

$67$ | |

$60$ | |

$58$ |

Question 3 Explanation:

The correct choice is (D). The degree of a polynomial is the power of its largest term. The largest term is $2x^7$. Since the power of this term is 7, $d$, is 7.

The term with no variable is the constant term. Here $c$, is 3.

Finally, the product of the non-constant coefficients of the polynomial, $p$, is $2 × 5 × 4 = 40$.

Plugging in $d$, $c$, and $p$ into $3d - c + p$, we get:

$3(7) - 3 + 40 $

$= 21 - 3 + 40$

$= 58$

The term with no variable is the constant term. Here $c$, is 3.

Finally, the product of the non-constant coefficients of the polynomial, $p$, is $2 × 5 × 4 = 40$.

Plugging in $d$, $c$, and $p$ into $3d - c + p$, we get:

$3(7) - 3 + 40 $

$= 21 - 3 + 40$

$= 58$

Question 4 |

### Which of the following is equivalent to $(8^x)^{3x} = 24$ where $x ≥ 0$?

$x = \dfrac{\log_8 24}{4}$ | |

$x = \dfrac{3}{4}$ | |

$x = \sqrt{\dfrac{\log_8 24}{3}}$ | |

$x = \sqrt{\dfrac{\log_3 24}{8}}$ |

Question 4 Explanation:

The correct choice is (C). The equation $(8^x)^{3x} = 24$ can be simplified to $8^{3x^2} = 24$.

We can rewrite this as:

$\require{cancel} \cancel{\log_8}(\cancel{8}^{3x^2}) = \log_8 (24)$

$3x^2 = \log_8 24$

Solving for $x$, we get: $x = \sqrt{\dfrac{\log_8 24}{3}}$

We can rewrite this as:

$\require{cancel} \cancel{\log_8}(\cancel{8}^{3x^2}) = \log_8 (24)$

$3x^2 = \log_8 24$

Solving for $x$, we get: $x = \sqrt{\dfrac{\log_8 24}{3}}$

Question 5 |

### Square $ABCD$ has an area of $16 \text{m}^2$. In sector $AE$, $AB ≅ EB$. What is the total perimeter of the figure above?

$2π + 20$ | |

$2π + 16$ | |

$8π + 16$ | |

$8π + 20$ |

Question 5 Explanation:

The correct choice is (B). Since square $ABCD$ has an area of $16 \text{m}^2$, it has a side length of $4\text{m}$. Therefore:

$AD + DC + CB + BE$ $= 4 × 4 = 16$

Next, we find the perimeter of the quarter-circle which is equivalent to:

$\dfrac{2π(4)}{4} = 2π$

Summing up, the perimeter of the entire figure is $2π + 16$.

$AD + DC + CB + BE$ $= 4 × 4 = 16$

Next, we find the perimeter of the quarter-circle which is equivalent to:

$\dfrac{2π(4)}{4} = 2π$

Summing up, the perimeter of the entire figure is $2π + 16$.

Question 6 |

### Which of the following choices is a valid factorization of the polynomial $28x^3 - 7x + 9$?

$7x(4x^2 - 1)$ | |

$7x(2x + 1)(2x - 1) + 9$ | |

$9x(4x^2 - 1)$ | |

$-7x(4x^2 - 1)(2x + 1)$ $(2x - 1) + 9$ |

Question 6 Explanation:

The correct choice is (B). We first factor out $7x$ from the first two terms of the polynomial $28x^3 - 7x + 9$. We get:

$7x(4x^2 - 1) + 9$

Next, we can factor out $4x^2 - 1$ into:

$(2x + 1)(2x - 1)$

Finally, we are left with our answer of $7x(2x + 1)(2x - 1) + 9$.

$7x(4x^2 - 1) + 9$

Next, we can factor out $4x^2 - 1$ into:

$(2x + 1)(2x - 1)$

Finally, we are left with our answer of $7x(2x + 1)(2x - 1) + 9$.

Question 7 |

### Movie tickets for the next Avengers movie cost from anywhere from \$15 to \$36. Each movie theater can fit no more than 200 adults and children. Let $c$ represent the cost of an individual ticket. Let $a$ represent the number of adults and $h$ represent the number of children in a given theater. Which system of inequalities best models the situation described above?

$15 ≤ c ≤ 36$
$a + h ≥ 200$ | |

$15 ≤ c ≤ 36$
$a + h > 200$ | |

$15 < c ≤ 36$
$a + h ≤ 200$ | |

$15 ≤ c ≤ 36$
$a + h ≤ 200$ |

Question 7 Explanation:

The correct choice is (D). The problem states that tickets are priced at \$15 or more. This means greater that $c$ is greater than or equal to 15. And they are \$36 or less, meaning that $c$ is less than or equal to 36:

$15 ≤ c ≤ 36$

Next, each movie theater can fit no more than 200 adults and children. In other words, it can fit 200 or less. This means that $a + h$ will be less than or equal to 200:

$a + h ≤ 200$

$15 ≤ c ≤ 36$

Next, each movie theater can fit no more than 200 adults and children. In other words, it can fit 200 or less. This means that $a + h$ will be less than or equal to 200:

$a + h ≤ 200$

Question 8 |

### What are the solutions to the following equation?

$(2x + \sqrt{2})(5x + \sqrt{19})$ $= 0$$x = \dfrac{-\sqrt{2}}{2}$ $\text{ and }$ $\dfrac{-\sqrt{19}}{5}$ | |

$x = \dfrac{\sqrt{2}}{2}$ $\text{ and }$ $\dfrac{\sqrt{19}}{5}$ | |

$x = \dfrac{-\sqrt{2}}{2}$ $\text{ and }$ $\dfrac{\sqrt{19}}{5}$ | |

$x = \dfrac{\sqrt{2}}{2}$ $\text{ and }$ $\dfrac{-\sqrt{19}}{5}$ |

Question 8 Explanation:

The correct choice is (A). Using the zero product property, we set $(2x + \sqrt{2})$ equal to $0$ and $(5x + \sqrt{19})$ equal to $0$.

Solving $2x + \sqrt{2} = 0$, we get: $x = \dfrac{-\sqrt{2}}{2}$

Solving $5x + \sqrt{19} = 0$, we get: $x = \dfrac{-\sqrt{19}}{5}$

Therefore, the zeroes of the equation are:

$x = \dfrac{-\sqrt{2}}{2}$ $\text{ and }$ $\dfrac{-\sqrt{19}}{5}$

Solving $2x + \sqrt{2} = 0$, we get: $x = \dfrac{-\sqrt{2}}{2}$

Solving $5x + \sqrt{19} = 0$, we get: $x = \dfrac{-\sqrt{19}}{5}$

Therefore, the zeroes of the equation are:

$x = \dfrac{-\sqrt{2}}{2}$ $\text{ and }$ $\dfrac{-\sqrt{19}}{5}$

Question 9 |

### Right triangle $ABC$ is shown above. Benedict uses the equation $\tan^{-1}(\frac{8}{15})$ to find the angle measure of $θ$. What is the value of $\sin θ + \cos θ$?

$\dfrac{23}{17}$ | |

$\dfrac{7}{15}$ | |

$\dfrac{8}{15}$ | |

$\dfrac{15}{17}$ |

Question 9 Explanation:

The correct choice is (A). Since the tangent of a triangle can be found by dividing the length of the side opposite to the angle $θ$ by the length of the side adjacent to the angle $θ$. The problem states that $\tan θ = \frac{8}{15}$, which means that $AB = 8$ and $BC = 15$.

The problem also states that the problem is a right triangle, which means that $AC = \sqrt{8^2 + 15^2} = 17$ via the Pythagorean theorem.

Next, we find:

$\sin θ =$ $\dfrac{ \text{Length of side opposite to } θ}{\text{Length of hypotenuse}}$ $= \dfrac{8}{17}$

$\cos θ =$ $\dfrac{\text{Length of side adjacent to } θ}{\text{Length of hypotenuse}}$ $= \dfrac{15}{17}$

Adding $\sin θ$ to $\cos θ$, we get: $\dfrac{8}{17} + \dfrac{15}{17} = \dfrac{23}{17}$

The problem also states that the problem is a right triangle, which means that $AC = \sqrt{8^2 + 15^2} = 17$ via the Pythagorean theorem.

Next, we find:

$\sin θ =$ $\dfrac{ \text{Length of side opposite to } θ}{\text{Length of hypotenuse}}$ $= \dfrac{8}{17}$

$\cos θ =$ $\dfrac{\text{Length of side adjacent to } θ}{\text{Length of hypotenuse}}$ $= \dfrac{15}{17}$

Adding $\sin θ$ to $\cos θ$, we get: $\dfrac{8}{17} + \dfrac{15}{17} = \dfrac{23}{17}$

Question 10 |

### Suppose $AC ≅ EC$. Which of the following must be true in order to prove that triangles $ABC$ and $EDC$ are congruent?

$AB ≅ ED$ | |

$∠ACB ≅ ∠ECD$ | |

$CD ≅ CE$ | |

$BC ≅ DC$ |

Question 10 Explanation:

The correct choice is (D). First, we know that $∠BCA ≅ ∠DCE$ since they are vertical angles. We are also given that $AC ≅ EC$.

Looking at choice (D), we see that if $BC ≅ DC$, triangles $ABC$ and $EDC$ would be congruent using the Side-Angle-Side postulate.

Looking at choice (D), we see that if $BC ≅ DC$, triangles $ABC$ and $EDC$ would be congruent using the Side-Angle-Side postulate.

Question 11 |

### If $x > 0$ and $y > 0$, which of the following expressions is equivalent to $\dfrac{\sqrt{x} + \sqrt{y}}{xy}$?

$\dfrac{\sqrt{x + y}}{xy}$ | |

$\dfrac{x + y}{\sqrt{x + y}}$ | |

$\dfrac{1}{y\sqrt{x}} + \dfrac{1}{x\sqrt{y}}$ | |

$\dfrac{1}{\sqrt{xy}}$ |

Question 11 Explanation:

The correct choice is (C).

The expression $\dfrac{\sqrt{x} + \sqrt{y}}{xy}$ can be separated into: $\dfrac{\sqrt{x}}{xy} + \dfrac{\sqrt{y}}{xy}$

This expression can be rewritten as:

$\dfrac{x^{\frac{1}{2}}}{xy} + \dfrac{y^{\frac{1}{2}}}{xy}$ $= \dfrac{1}{y\sqrt{x}} + \dfrac{1}{x\sqrt{y}}$

The expression $\dfrac{\sqrt{x} + \sqrt{y}}{xy}$ can be separated into: $\dfrac{\sqrt{x}}{xy} + \dfrac{\sqrt{y}}{xy}$

This expression can be rewritten as:

$\dfrac{x^{\frac{1}{2}}}{xy} + \dfrac{y^{\frac{1}{2}}}{xy}$ $= \dfrac{1}{y\sqrt{x}} + \dfrac{1}{x\sqrt{y}}$

Question 12 |

### Which of the following equations has a slope parallel to $y = \dfrac{1}{2}(x + 3)$ and passes through the point $(3, 16)$?

$y = -2x + \dfrac{29}{2}$ | |

$y = 2x + \dfrac{3}{2}$ | |

$y = \dfrac{1}{2}x + \dfrac{29}{2}$ | |

$y = -\dfrac{1}{2}x + \dfrac{3}{2}$ |

Question 12 Explanation:

The correct choice is (C). A line parallel to one with the equation of $y = \frac{1}{2}(x + 3)$ has a slope of $\frac{1}{2}$.

Applying the point-slope formula to find the equation of the line that passes through $(3, 16)$, we get $y - 16 = \frac{1}{2}(x - 3)$, which can be simplified to $y = \frac{1}{2}x + \frac{29}{2}$.

Applying the point-slope formula to find the equation of the line that passes through $(3, 16)$, we get $y - 16 = \frac{1}{2}(x - 3)$, which can be simplified to $y = \frac{1}{2}x + \frac{29}{2}$.

Question 13 |

### Which of the following are the solutions to $3x^2 + 9x + 12 = 0$?

*Express $\sqrt{-1}$ as i*.

$- \dfrac{3}{2} ± \dfrac{i\sqrt{7}}{2}$ | |

$\dfrac{3}{2} ± \dfrac{3i\sqrt{7}}{2}$ | |

$- \dfrac{3}{2} ± \dfrac{\sqrt{51}}{6}$ | |

$\dfrac{3}{2} ± \dfrac{-i\sqrt{23}}{2}$ |

Question 13 Explanation:

The correct choice is (A). We can apply the quadratic formula to $3x^2 + 9x + 12 = 0$.

Plugging in 3, 9, and 12 for $a$,$b$, and $c$ respectively into:

$\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$, we get:

$\dfrac{-9 ± \sqrt{-63}}{6}$ $= \dfrac{-9 ± \sqrt{-1 × 9 × 7}}{6}$

$= \dfrac{-9 ± \sqrt{9 × -1 × 7}}{6}$ $= \dfrac{-9 ± \sqrt{9} \sqrt{-1} \sqrt{7}}{6}$

$= \dfrac{-9 ± 3i \sqrt{7}}{6}$ $= -\dfrac{9}{6} ± \dfrac{3i\sqrt{7}}{2}$

$= -\dfrac{3}{2} ± \dfrac{i\sqrt{7}}{2}$

Plugging in 3, 9, and 12 for $a$,$b$, and $c$ respectively into:

$\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$, we get:

$\dfrac{-9 ± \sqrt{-63}}{6}$ $= \dfrac{-9 ± \sqrt{-1 × 9 × 7}}{6}$

$= \dfrac{-9 ± \sqrt{9 × -1 × 7}}{6}$ $= \dfrac{-9 ± \sqrt{9} \sqrt{-1} \sqrt{7}}{6}$

$= \dfrac{-9 ± 3i \sqrt{7}}{6}$ $= -\dfrac{9}{6} ± \dfrac{3i\sqrt{7}}{2}$

$= -\dfrac{3}{2} ± \dfrac{i\sqrt{7}}{2}$

Question 14 |

### The function above shows the value of a used bike over time. Which of the following is true about the function?

$\text{The}$ $\text{initial}$ $\text{value}$ $\text{of}$ $\text{the}$ $\text{bike}$ $\text{is}$ $\$300$ | |

$\text{The}$ $\text{value}$ $\text{of}$ $\text{the}$ $\text{bike}$ $\text{approaches}$ $\$0$ $\text{as}$ $\text{time}$ $\text{progresses.}$ | |

$\text{The}$ $\text{value}$ $\text{of}$ $\text{the}$ $\text{bike}$ $\text{is}$ $\text{stagnant}$ $\text{after}$ $\text{year}$ $5.$ | |

$\text{The}$ $\text{total}$ $\text{value}$ $\text{of}$ $\text{the}$ $\text{bike}$ $\text{is}$ $15$ $\text{thousand}$ $\text{dollars.}$ |

Question 14 Explanation:

The correct choice is (B). Based on the graph, we see that initially (

*t*= 0) the bike has a value of \$3,000. After 5 years (*t*= 5), the bike's value is approximately \$1,000. As time progresses, the bike’s value approaches \$0.Question 15 |

### Which of the following is true of the function shown below?

$f(x) =$ $-0.7x^3 + 3.14x^2$ $ +\, 6.28x + 121$$\text{It}$ $\text{has}$ $\text{a}$ $\text{degree}$ $\text{of}$ $6,$ $\text{and}$ $\text{as}$ $x$ $\text{approaches}$ $∞,$ $y$ $\text{approaches}$ $-∞.$ | |

$\text{It}$ $\text{has}$ $\text{an}$ $\text{even}$ $\text{degree},$ $\text{and}$ $\text{as}$ $y$ $\text{approaches}$ $∞,$ $x$ $\text{approaches}$ $-∞.$ | |

$\text{It}$ $\text{has}$ $\text{a}$ $\text{degree}$ $\text{of}$ $3,$ $\text{and}$ $\text{as}$ $x$ $\text{approaches}$ $∞,$ $y$ $\text{approaches}$ $-∞.$ | |

$\text{It}$ $\text{has}$ $\text{a}$ $\text{prime}$ $\text{degree},$ $\text{and}$ $\text{as}$ $x$ $\text{approaches}$ $∞,$ $y$ $\text{approaches}$ $∞.$ |

Question 15 Explanation:

The correct choice is (C). Examining the function above, we know that its degree is odd and it has a negative leading coefficient.

In functions that have odd degrees and negative leading coefficients, as $x$ approaches ∞, $y$ approaches -∞, and as $x$ approaches -∞, $y$ approaches ∞. Thus, the only choice that correctly identifies the degree and the end behavior of the graph is choice C.

In functions that have odd degrees and negative leading coefficients, as $x$ approaches ∞, $y$ approaches -∞, and as $x$ approaches -∞, $y$ approaches ∞. Thus, the only choice that correctly identifies the degree and the end behavior of the graph is choice C.

Question 16 |

### Which of the following are true about linear equations of form $y = ax + b, a ≠ 0$?

$\text{A}$ $\text{quadratic}$ $\text{equation}$ $\text{is}$ $\text{linear.}$ | |

$\text{Such}$ $\text{linear}$ $\text{equations}$ $\text{can}$ $\text{produce}$ $\text{vertical}$ $\text{or}$ $\text{horizontal}$ $\text{lines.}$ | |

$\text{Exponential}$ $\text{functions}$ $\text{are}$ $\text{linear.}$ | |

$\text{Linear}$ $\text{equations}$ $\text{are}$ $\text{one}$ $\text{to}$ $\text{one}$ $\text{functions.}$ |

Question 16 Explanation:

The correct choice is (D). The only statement true about a linear equation is that it is a type of one-to-one function. A function is one-to-one if no two elements in the domain of the function correspond to the same element in the range of the function.

Graphically, a one-to-one function passes the horizontal line test. Since a line passes the horizontal line test, linear equations are one-to-one functions.

Graphically, a one-to-one function passes the horizontal line test. Since a line passes the horizontal line test, linear equations are one-to-one functions.

Question 17 |

### Relation $z$ is pictured above. What is the value of $z(z(-8))$?

$8$ | |

$4$ | |

$2$ | |

$-2$ |

Question 17 Explanation:

The correct choice is (C). First, we find the value of $z(-8)$ which is $4$. Next, we plug $4$ back into the function. We are left to find $z(4)$ which is equal to $2$.

Question 18 |

### Consider triangle $ABC$ above. Which of the following could be the angle measure of $∠B$?

$133.224°$ | |

$46.776°$ | |

$127.228°$ | |

$80.224°$ |

Question 18 Explanation:

The correct choice is (D). Applying the law of sines to triangle $ABC$, we get:

$\dfrac{\sin 53}{13.7} = \dfrac{\sin C}{12.5}$

Cross multiplying and dividing, we get:

$\dfrac{\sin 53 × 12.5}{13.7} = \sinC$

Taking the inverse sin of both sides, we get: $C = 46.776°$

Finding angle $B$, we get: $180 - 53 - 46.776$ $= 80.224°$

(Note that the inverse sin actually gives two answers:

$\dfrac{\sin 53}{13.7} = \dfrac{\sin C}{12.5}$

Cross multiplying and dividing, we get:

$\dfrac{\sin 53 × 12.5}{13.7} = \sinC$

Taking the inverse sin of both sides, we get: $C = 46.776°$

Finding angle $B$, we get: $180 - 53 - 46.776$ $= 80.224°$

(Note that the inverse sin actually gives two answers:

*C*= 46.776° and 133.224°. However, only one of these is possible with the given triangle, since 133.224° + 53° > 180°.)Question 19 |

### Which of the following situations could the function above represent?

$\text{The}$ $\text{cost}$ $\text{of}$ $\text{oranges}$ $\text{at}$ $\text{a}$ $\text{unit}$ $\text{price}$ $\text{of}$ $\$2.55$ $\text{per}$ $\text{pound.}$ | |

$\text{The}$ $\text{total}$ $\text{taxi}$ $\text{fare}$ $\text{that}$ $\text{has}$ $\text{a}$ $\text{rate}$ $\text{of}$ $\$4.55$ $\text{per}$ $\text{half}$ $\text{mile.}$ | |

$\text{The}$ $\text{area}$ $\text{of}$ $\text{a}$ $\text{circle}$ $\text{in}$ $\text{relation}$ $\text{to}$ $\text{its}$ $\text{radius.}$ | |

$\text{The}$ $\text{value}$ $\text{of}$ $\text{a}$ $\text{car}$ $\text{over}$ $\text{time.}$ |

Question 19 Explanation:

The correct choice is (C). Since the relationship that is graphed is quadratic where $0 ≤ x ≤ ∞$, we see that the only choice that represents a quadratic relationship is a circle's area as a function of its radius.

Question 20 |

### A quadratic function has roots at $(-3, 0)$ and $(5, 0)$. Which of the following could represent the minimum point of the function?

$(1, -16)$ | |

$(1.5, -16)$ | |

$(2, -16)$ | |

$(1, 0)$ |

Question 20 Explanation:

The correct choice is (A). We are given that the two zeroes of the graph are $(-3, 0)$ and $(5, 0)$, which means that the minimum point of the function has an $x$ value of $\dfrac{-3 + 5}{2} = 1$.

The equation of this parabola is $y = (x + 3)(x - 5)$ from the given roots. This means the $y$-value of the parabola at $x = 1$ is $(1 + 3)(1 - 5) = -16$. Therefore, the correct answer is $(1, -16)$.

The equation of this parabola is $y = (x + 3)(x - 5)$ from the given roots. This means the $y$-value of the parabola at $x = 1$ is $(1 + 3)(1 - 5) = -16$. Therefore, the correct answer is $(1, -16)$.

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 20 questions to complete.

List |