Graphing Quadratics and Inequalities: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Multiple choice

20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.

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Question 1 of 20

1

The graph of $y = -3x^2 + 5x - 1$ is a parabola. How does it open, and how does its width compare with the graph of $y = x^2$ ?

Answer choices for question 1 A It opens downward and is wider than $y = x^2$ . B It opens upward and is narrower than $y = x^2$ . C It opens downward and is narrower than $y = x^2$ . D It opens upward and is wider than $y = x^2$ .
2

Which is the standard equation of the circle with center $(4, -2)$ and radius $5$ ?

Answer choices for question 2 A $(x - 4)^2 + (y + 2)^2 = 25$ B $(x - 4)^2 + (y - 2)^2 = 25$ C $(x - 4)^2 + (y + 2)^2 = 5$ D $(x + 4)^2 + (y - 2)^2 = 25$
3

What is the vertex of the parabola $y = -2(x + 3)^2 + 7$ ?

Answer choices for question 3 A $(3, 7)$ B $(3, -7)$ C $(-3, -7)$ D $(-3, 7)$
4

Advanced. This question goes beyond core Algebra I. It is not required by the course.

For which values of $x$ is $(x + 1)(x - 4) > 0$ ?

Answer choices for question 4 A $-1 < x < 4$ B $x < -1$ or $x > 4$ C $x > -1$ or $x < 4$ D $x < -4$ or $x > 1$
5

What is the distance between the points $(-2, 3)$ and $(4, -5)$ ?

Answer choices for question 5 A $10$ B $14$ C $\sqrt{28}$ D $100$
6

The quantity $y = x^2 - 8x + 10$ has a minimum. What is the minimum value?

Answer choices for question 6 A $4$ B $10$ C $-6$ D $58$
7

How many times does the graph of $y = x^2 + 4x + 9$ cross the $x$ -axis?

Answer choices for question 7 A Twice. B It does not cross the $x$ -axis. C Once. D It cannot be found without graphing.
8

What are the center and radius of the circle $(x - 5)^2 + (y + 1)^2 = 49$ ?

Answer choices for question 8 A Center $(-5, 1)$ , radius $7$ B Center $(5, -1)$ , radius $49$ C Center $(-5, 1)$ , radius $49$ D Center $(5, -1)$ , radius $7$
9

The circle $(x - 1)^2 + (y - 2)^2 = 169$ is given. Where does the point $(6, 14)$ lie relative to it?

Answer choices for question 9 A Inside the circle. B Outside the circle. C On the circle. D Its position cannot be decided from the equation alone.
10

Advanced. This question goes beyond core Algebra I. It is not required by the course.

What is the solution of $-x^2 + 2x + 8 \ge 0$ ?

Answer choices for question 10 A $x \le -2$ or $x \ge 4$ B $-2 \le x \le 4$ C $-2 < x < 4$ D $-4 \le x \le 2$
11

The parabola $y = 2x^2 + 12x + 13$ has a vertex. What are its coordinates?

Answer choices for question 11 A $(-3, -5)$ B $(3, 67)$ C $(-3, 13)$ D $(-3, 31)$
12

What is the radius of the circle $(x - 3)^2 + (y + 6)^2 = 20$ ?

Answer choices for question 12 A $20$ B $10$ C $4\sqrt{5}$ D $2\sqrt{5}$
13

Two numbers add to $20$ . What is the greatest possible value of their product?

Answer choices for question 13 A $10$ B $100$ C $400$ D $20$
14

A circle has equation $(x + 2)^2 + (y - 1)^2 = 16$ . Does the point $(1, 5)$ lie inside, on, or outside the circle?

Answer choices for question 14 A Inside the circle. B On the circle. C Outside the circle. D There is not enough information to decide.
15

What is the center of the circle $2x^2 + 2y^2 - 8x + 12y - 6 = 0$ ?

Answer choices for question 15 A $(2, -3)$ B $(-2, 3)$ C $(4, -6)$ D $(-4, 6)$
16

The parabola $y = 3x^2 - 6x + 4$ opens upward. How many times does it cross the $x$ -axis?

Answer choices for question 16 A Twice. B Once. C The count depends on the graph and cannot be found from the equation. D It never crosses the $x$ -axis.
17

Advanced. This question goes beyond core Algebra I. It is not required by the course.

A solution to $x^2 - x - 6 > 0$ is written as $-2 < x < 3$ . Which single misreading produces that answer?

Answer choices for question 17 A The interval between the roots was chosen, where the parabola is below the axis, instead of the two rays outside them. B The inequality symbol was reversed while multiplying by a negative. C The quadratic was factored with the wrong roots. D Strict endpoints were used where inclusive ones belong.
18

Rewriting $x^2 + y^2 + 6x - 4y - 12 = 0$ in standard form, what are its center and radius?

Answer choices for question 18 A Center $(3, -2)$ , radius $5$ B Center $(-3, 2)$ , radius $5$ C Center $(-3, 2)$ , radius $25$ D Center $(-3, 2)$ , radius $\sqrt{12}$
19

Advanced. This question goes beyond core Algebra I. It is not required by the course.

What is the complete solution set of $(x + 5)^2 > 0$ ?

Answer choices for question 19 A All real numbers. B $x > -5$ . C No solution. D Every real number except $x = -5$ .
20

A quantity is modeled by $y = -4(x - 5)^2 + 90$ . What is the maximum value of $y$ , and at what input does it occur?

Answer choices for question 20 A Maximum value $5$ , occurring at $x = 90$ . B Maximum value $5$ , occurring at $x = 5$ . C Maximum value $90$ , occurring at $x = 5$ . D There is no maximum; the quantity grows without bound.

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Where you stand

Read a parabola's direction and relative width from the sign and size of $a$ , its axis of symmetry $x=-\tfrac{b}{2a}$ , and its $y$ -intercept $(0,c)$ . Review: Parabolas
Locate the vertex as a point, its $x$ from $-\tfrac{b}{2a}$ and its height by substituting back, and read $(h,k)$ from vertex form $y=a(x-h)^2+k$ with the sign flip. Review: Parabolas
Use the discriminant $b^2-4ac$ to count a parabola's real $x$ -intercepts, and to decide the all-reals, no-solution, or repeated-root boundary of a quadratic inequality. Review: Parabolas , Quadratic Inequalities
Write a circle's standard equation $(x-h)^2+(y-k)^2=r^2$ from its center and radius, and read the center and radius back, handling the sign trap and the radius trap $r=\sqrt{\text{right side}}$ . Review: The Equation of a Circle
Apply the distance formula, find a circle's center as the midpoint of a diameter, and decide whether a point lies inside, on, or outside a circle by comparing its squared distance from the center with $r^2$ . Review: The Equation of a Circle
Convert a circle from general form to standard form, dividing out a common leading coefficient and completing the square on each variable, and classify the result as a circle, a single point, or empty from the sign of the right side. Review: The Equation of a Circle
Solve a quadratic inequality by comparing to zero, finding roots, and reading the sign on each interval (outside/between for an upward parabola, reversed for a downward one), matching strict or inclusive endpoints and writing an outside solution as two rays. Review: Quadratic Inequalities
Use the sign of $a$ to tell a maximum from a minimum, and distinguish the optimizing input $x=-\tfrac{b}{2a}$ from the optimal value found by substituting it back, reporting the one a question asks for. Review: Quadratic Optimization

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two equal heights

The graph of $y=x^2+bx+c$ passes through $(-1,3)$ and $(5,3)$ . Find $b$ and $c$ .
1. Hint 1
  Two inputs that give the same height sit the same distance either side of the axis of symmetry.
2. Hint 2
  Locate that axis, put it into $x=-\frac b{2a}$ to get $b$ , then substitute one of the points to get $c$ .
Answer

$b=-4$ ; $c=-2$ .

Full solution

The two given points share the height $3$ , so they are mirror images across the axis of symmetry, which lies halfway between them.
$x=\frac{-1+5}{2}$

This is $x=2$ .
Here $a=1$ , so the axis formula requires
$-\frac b2=2$

Multiplying by $-2$ gives $b=-4$ .
Substituting the point $(-1,3)$ into $y=x^2-4x+c$ gives
$1+4+c=3$

Hence $c=-2$ , and the y-intercept is $(0,-2)$ .
The other point checks the pair, since $25-20-2=3$ .

Answer

$b=-4$ ; $c=-2$ .

Key idea

Two points at the same height place the axis of symmetry, which then fixes the linear coefficient.
Problem 1: Mark complete
Problem 2 A circular outline

The figure shows a circle. Write its equation in standard form.

A circle drawn on the coordinate plane.

Text description of this figure

A square coordinate grid with equal unit spacing on both axes. The horizontal x-axis is numbered at every whole number from negative two to six and the vertical y-axis at every whole number from negative five to three, with tick marks and gridlines at every whole number and the origin labeled 0. One circle is drawn on the grid as a plain unshaded outline. Its leftmost point sits on the vertical gridline at negative one, its rightmost point on the vertical gridline at five, its highest point on the horizontal gridline at two, and its lowest point on the horizontal gridline at negative four. Nothing else is drawn: no center point, no radius segment, no labeled points and no equation.
1. Hint 1
  Read the horizontal and vertical extremes to locate the middle of the circle.
2. Hint 2
  The radius is the distance from that middle to any extreme; square it on the right side.
Answer

$(x-2)^2+(y+1)^2=9$ .

Full solution

The circle extends horizontally from $-1$ to $5$ and vertically from $-4$ to $2$ .
Opposite extremes straddle the center, so each center coordinate is the midpoint of one pair.
$\left(\frac{-1+5}{2},\frac{-4+2}{2}\right)=(2,-1)$

Half of each span is the radius.
$r=\frac{5-(-1)}{2}$

This is $3$ , and the vertical span $2-(-4)$ gives the same half, $3$ .
Standard form therefore gives
$(x-2)^2+(y+1)^2=9$

Stepping $3$ from $(2,-1)$ lands on $(-1,-1)$ , $(5,-1)$ , $(2,2)$ and $(2,-4)$ , the four extremes drawn on the grid.

Answer

$(x-2)^2+(y+1)^2=9$ .

Key idea

A circle's horizontal and vertical extremes straddle its center, so their midpoints give the center and half their span gives the radius.
Problem 2: Mark complete
Problem 3 A target peak

For real $x$ , the greatest value of $Q(x)=-x^2+8x+c$ must be $20$ . Find $c$ .
1. Hint 1
  Locate the maximizing input from the variable coefficients.
2. Hint 2
  The value at that input supplies an equation for the unknown constant.
Answer

$c=4$ .

Full solution

The leading coefficient is negative, so the vertex gives the greatest value.
It occurs at
$x=-\frac8{2(-1)}$

This is $x=4$ .
The required output gives
$-16+32+c=20$

Thus $c=4$ .
The resulting form $Q(x)=-(x-4)^2+20$ confirms that the greatest value is exactly $20$ .

Answer

$c=4$ .

Key idea

A required optimal value can determine a constant after the optimizing input is found.
Problem 3: Mark complete
Problem 4 A diameter and a point

A circle has diameter endpoints $A=(3,-4)$ and $B=(5,2)$ . Write its equation in standard form and decide whether $P=(6,0)$ lies inside, on, or outside the circle.
1. Hint 1
  The center is halfway between the diameter endpoints.
2. Hint 2
  Find the radius squared from either endpoint, then compare the squared distance to $P$ .
Answer

$(x-4)^2+(y+1)^2=10$ ; $P$ is inside the circle.

Full solution

Averaging the endpoint coordinates gives center $(4,-1)$ .
The squared radius is
$r^2=(3-4)^2+(-4+1)^2$

This is $10$ , giving
$(x-4)^2+(y+1)^2=10$

The squared distance from the center to $P$ is
$d^2=(6-4)^2+(0+1)^2$

This is $5<10$ , so $P$ is inside.
The endpoint $B$ gives $(5-4)^2+(2+1)^2=10$ as well, so both ends of the diameter lie on the circle.

Answer

$(x-4)^2+(y+1)^2=10$ ; $P$ is inside the circle.

Key idea

Diameter data determine the circle, and a squared-distance comparison locates another point relative to it.
Problem 4: Mark complete
Problem 5 A combined expression

Find the vertex of $y=(2x-3)^2-4x+5$ and its least value when $3\le x\le5$ .
1. Hint 1
  Expand to identify the leading and linear coefficients.
2. Hint 2
  Find the vertex as a point, then check whether its input is one of the permitted values.
Answer

Vertex $(2,-2)$ ; the least value for $3\le x\le5$ is $2$ , reached at $x=3$ .

Full solution

Expanding gives
$y=4x^2-16x+14$

The vertex input is
$x=-\frac{-16}{2(4)}$

This is $2$ .
Substitution gives $y=16-32+14=-2$ , so the vertex is $(2,-2)$ , and the form $y=4(x-2)^2-2$ confirms it.

The input $2$ is not permitted here.
The leading coefficient is positive, so the height climbs as $x$ moves right of $2$ , and the least permitted value is at the left endpoint.
$y=4(9)-16(3)+14$

This is $2$ .
The right endpoint gives $y=100-80+14=34$ , which is larger, as expected.

Answer

Vertex $(2,-2)$ ; the least value for $3\le x\le5$ is $2$ , reached at $x=3$ .

Key idea

When the vertex input is not permitted, the best value on an interval sits at the permitted input closest to it.
Problem 5: Mark complete
Problem 6 Two parameter settings

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Find the complete real solution set of $x^2-2kx+4>0$ for $k=1$ and for $k=3$ .
1. Hint 1
  The two values of $k$ can give different numbers of real roots.
2. Hint 2
  Check the discriminant in each case; if there are two roots, determine the sign on their three intervals.
Answer

$k=1$ : all real $x$ . $k=3$ : $x<3-\sqrt5$ or $x>3+\sqrt5$ .

Full solution

At $k=1$ , the quadratic is $x^2-2x+4$ .
Its discriminant is
$4-16=-12<0.$
It opens upward and has no real zeros, so it is positive for every real input.

At $k=3$ , the quadratic is $x^2-6x+4$ , with discriminant
$36-16=20>0.$
Its roots are
$x=3\pm\sqrt5$

The upward quadratic is positive outside these roots.
For example, inputs $0$ and $6$ give $4>0$ , while the middle input $3$ gives $-5<0$ .
Exclude both roots because the comparison is strict.

Answer

$k=1$ : all real $x$ . $k=3$ : $x<3-\sqrt5$ or $x>3+\sqrt5$ .

Key idea

The discriminant says whether a quadratic needs sign intervals, keeps one sign everywhere, or keeps one sign apart from a single repeated root.
Problem 6: Mark complete
Problem 7 Two distance records

Find every real point $(x,y)$ whose squared distance from $(-1,0)$ plus its squared distance from $(3,4)$ is $16$ . Describe the resulting graph.
1. Hint 1
  Write each squared distance in coordinate form and add the expressions.
2. Hint 2
  Expand, divide out the common leading coefficient, and complete both squares.
Answer

The single point $(1,2)$ .

Full solution

Write the two squared distances as $D_1=(x+1)^2+y^2$ and
$D_2=(x-3)^2+(y-4)^2.$
The condition is
$D_1+D_2=16$

Expanding the four squares and collecting like terms gives
$2x^2+2y^2-4x-8y+26=16$

Subtract $16$ from both sides, then divide every term by $2$ .
$x^2+y^2-2x-4y+5=0$

Complete the two squares.
$(x-1)^2+(y-2)^2=0$

Both real squares must vanish, so the graph is the single point $(1,2)$ .
Its squared distances to the given points are $8$ and $8$ , whose sum is $16$ .

Answer

The single point $(1,2)$ .

Key idea

A sum of squared distances can reduce to a circle-form equation that represents just one point.
Problem 7: Mark complete
Problem 8 A student's intercept count

A student decides how many times the graph of $y=-2x^2+5x-4$ crosses the x-axis. The student computes $5^2-4(-2)(-4)=25+32=57$ , and concludes from its positive sign that there are two x-intercepts. Find the student's error, give the corrected value, and state the true number of x-intercepts.
1. Hint 1
  How many times a parabola meets the x-axis is settled by the sign of $b^2-4ac$ , never by its size.
2. Hint 2
  Read the coefficients with their signs, and evaluate $-4ac$ as one signed product before adding it to $b^2$ .
Answer

The term $-4ac$ is $-32$ , not $+32$ ; the corrected value is $-7$ ; the graph has no x-intercepts.

Full solution

The coefficients are $a=-2$ , $b=5$ and $c=-4$ , so the term added to $b^2$ is
$-4ac=-4(-2)(-4)$

The two negative factors give $+8$ , and the leading $-4$ turns that into $-32$ .
The student kept it positive, which is the error.

The corrected value is therefore
$5^2-32=-7$

A negative value means $-2x^2+5x-4=0$ has no real solution, so the graph has no x-intercepts.
Its vertex is at $x=\frac54$ , with height $-\frac78$ , and a downward parabola whose highest point lies below the axis never reaches it.

Answer

The term $-4ac$ is $-32$ , not $+32$ ; the corrected value is $-7$ ; the graph has no x-intercepts.

Key idea

The sign of $b^2-4ac$ counts a parabola's x-intercepts, so every sign inside the product $-4ac$ has to be carried through.
Problem 8: Mark complete
Problem 9 A signal interval

Advanced. This question goes beyond core Algebra I. It is not required by the course.

A signal has strength $S$ microvolts at time $t$ seconds, where $S=12t-t^2$ for $0\le t\le12$ . Find its greatest strength, the time at which that occurs, and every time at which the strength is at least half that greatest value.
1. Hint 1
  Find the highest point of the strength quadratic first.
2. Hint 2
  Use half that value as the right side of an inequality.
3. Hint 3
  Solve the inequality and check that its time interval stays within the stated model domain.
Answer

Greatest strength $36$ microvolts at $t=6$ seconds; at least half strength for $6-3\sqrt2\le t\le6+3\sqrt2$ seconds.

Full solution

The leading coefficient is negative, so the greatest value occurs at
$t=-\frac{12}{2(-1)}$

This is $6$ seconds, giving strength $36$ microvolts.
Half of that is $18$ microvolts.
The condition becomes
$t^2-12t+18\le0$

The two roots are
$t=6\pm3\sqrt2$

The upward quadratic is zero or negative between these roots, with both endpoints included.
Since $3\sqrt2<6$ , the entire interval lies within
$0\le t\le12.$
At its midpoint $t=6$ , the signal has its full greatest strength.

Answer

Greatest strength $36$ microvolts at $t=6$ seconds; at least half strength for $6-3\sqrt2\le t\le6+3\sqrt2$ seconds.

Key idea

An optimal value can supply the threshold for a second quadratic inequality.
Problem 9: Mark complete
Problem 10 Two plotted points

The figure marks $A=(-2,4)$ and $B=(2,4)$ . A student says that a parabola through both points must have no x-intercepts. Decide whether the student is correct, and justify your decision with a specific quadratic or a general argument.

The marked points $A$ and $B$ .

Text description of this figure

A square coordinate grid with equal unit spacing on both axes. The horizontal x-axis is numbered at every whole number from negative three to three and the vertical y-axis at every whole number from negative five to five, with tick marks and gridlines at every whole number and the origin labeled 0. Two points are marked with filled dots, each labeled above the dot with its name and coordinates: point A, two units left of the vertical axis and four units above the horizontal axis, and point B, two units right of the vertical axis and four units above the horizontal axis. Both points sit on the same horizontal gridline. No curve, line of symmetry, vertex or axis crossing is drawn.
1. Hint 1
  Two points do not determine all three coefficients of a quadratic.
2. Hint 2
  For a symmetric choice $y=ax^2+c$ , impose the point condition and choose coefficients whose discriminant is not negative.
Answer

False; for example, $y=2x^2-4$ has two x-intercepts.

Full solution

Choose
$y=2x^2-4$

At each of $x=-2$ and $x=2$ , its height is $4$ , so it passes through both marked points.
Its discriminant is
$D=0-4(2)(-4)$

This is $32>0$ , so it has two x-intercepts, at $x=-\sqrt2$ and $x=\sqrt2$ .
The student is therefore wrong, because two points above the axis do not establish a negative discriminant.

Answer

False; for example, $y=2x^2-4$ has two x-intercepts.

Key idea

A few points above the axis do not determine whether an entire parabola meets the axis.
Problem 10: Mark complete

Star problems Ten optional challenges to stretch your reasoning Advanced. This problem set goes beyond core Algebra I. You can skip it. → Continue to Chapter 11: Functions →

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