Grade 10 Mathematics: Quadratic Functions, Radical Equations, and Circles

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Vocabulary practice flashcards covering quadratic equations and functions, projectile motion and optimization models, radical equations, and the center-radius and general forms of circles.

50 Terms

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Monic Quadratic Equation with Zeros 22 and −5-5

The quadratic equation f(x)=x2+3x−10f(x) = x^2 + 3x - 10, formed by expanding (x−2)(x+5)(x - 2)(x + 5) where the leading coefficient is 11.

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Quadratic Equation with Zeros −2-2 and 33, and f(0)=12f(0) = 12

The function f(x)=−2(x+2)(x−3)f(x) = -2(x + 2)(x - 3), derived using factored form f(x)=a(x−r1)(x−r2)f(x) = a(x - r_1)(x - r_2) and solving for a=−2a = -2.

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Linear Factor for Zeros −4-4 and 11

An expression such as (x+4)(x + 4) or (x−1)(x - 1) that must divide the quadratic equation with roots −4-4 and 11.

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Quadratic Function Passing Through Points (−1,6)(-1, 6), (0,3)(0, 3), and (1,2)(1, 2)

The function f(x)=x2−2x+3f(x) = x^2 - 2x + 3, which satisfies all three coordinate points.

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yy-Intercept Constant (cc)

The constant term cc in f(x)=ax2+bx+cf(x) = ax^2 + bx + c, which equals −4-4 when the table contains (0,−4)(0, -4).

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Quadratic Equation with Vertex (1,−2)(1, -2) Passing Through (3,6)(3, 6)

The quadratic function f(x)=2(x−1)2−2f(x) = 2(x - 1)^2 - 2, determined using vertex form.

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Parabola with Zeros −3-3 and 22 Passing Through (0,−12)(0, -12)

The quadratic equation f(x)=2(x+3)(x−2)f(x) = 2(x + 3)(x - 2), derived using factored form with a=2a = 2.

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Vertex Form of a Quadratic Function

The form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k, which is most convenient when a graph provides its vertex (h,k)(h, k) and one other point.

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Area Function of a Garden Enclosed by 80 m80\,m of Fencing

The quadratic function A(x)=x(40−x)A(x) = x(40 - x), where xx is the length of one side and (40−x)(40 - x) is the adjacent side.

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Dimensions for Maximum Area with Perimeter 60 m60\,m

15 m15\,m by 15 m15\,m, forming a square to maximize the rectangular area.

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Greatest Possible Area for A(x)=−x2+36xA(x) = -x^2 + 36x

324 square units324\,\text{square units}, which occurs at the vertex where x=18x = 18.

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Time of Maximum Height for h(t)=−5t2+40t+2h(t) = -5t^2 + 40t + 2

4 s4\,s, calculated by evaluating the vertex time t=−b2a=−402(−5)t = -\frac{b}{2a} = -\frac{40}{2(-5)}.

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Maximum Height for h(t)=−4t2+24th(t) = -4t^2 + 24t

36 m36\,m, reached at time t=3 st = 3\,s.

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Profit Model P(x)P(x) for (50+x)(50 + x) Items at (100−2x)(100 - 2x) Pesos

The quadratic model P(x)=−2x2+5000P(x) = -2x^2 + 5000, calculated by expanding (50+x)(100−2x)(50 + x)(100 - 2x).

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Optimal Production Level for P(x)=−2x2+80x+1000P(x) = -2x^2 + 80x + 1000

x=20x = 20, the production value that maximizes profit at the vertex x=−802(−2)x = -\frac{80}{2(-2)}.

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Vertex yy-Coordinate in Profit Model (30,4500)(30, 4500)

The maximum daily profit of P4,500\text{P}4,500, corresponding to producing 3030 additional units.

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Vertex Coordinates in Projectile Height Model (2,18)(2, 18)

The point indicating that the object reaches its maximum height of 18 m18\,m after 2 s2\,s.

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Solution to x+7=5\sqrt{x + 7} = 5

x=18x = 18, obtained by squaring both sides to get x+7=25x + 7 = 25.

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Solution to 2x+1=82\sqrt{x + 1} = 8

x=15x = 15, determined by isolating the radical to x+1=4\sqrt{x + 1} = 4 and solving x+1=16x + 1 = 16.

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Solution to x+5=x−1\sqrt{x + 5} = x - 1

x=4x = 4, which satisfies the equation, whereas x=−1x = -1 is extraneous.

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Solvability of x+4=−3\sqrt{x + 4} = -3

It has no real solution because a principal square root cannot equal a negative number.

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General Form of (x−2)2+(y+1)2=16(x - 2)^2 + (y + 1)^2 = 16

x2+y2−4x+2y−11=0x^2 + y^2 - 4x + 2y - 11 = 0, found by expanding and collecting all terms on one side.

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Constant FF in the General Form of (x+3)2+(y−4)2=25(x + 3)^2 + (y - 4)^2 = 25

00, calculated from 32+(−4)2−25=9+16−253^2 + (-4)^2 - 25 = 9 + 16 - 25.

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Center-Radius Form of x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0

(x−3)2+(y+2)2=25(x - 3)^2 + (y + 2)^2 = 25, derived by completing the squares for xx and yy.

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Center-Radius Form of x2+y2+2x−8y+8=0x^2 + y^2 + 2x - 8y + 8 = 0

(x+1)2+(y−4)2=9(x + 1)^2 + (y - 4)^2 = 9, determined by completing the squares for xx and yy.

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Monic Quadratic with Zeros 22 and −5-5

The quadratic function f(x)=x2+3x−10f(x) = x^2 + 3x - 10, formed by expanding (x−2)(x+5)(x - 2)(x + 5).

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Quadratic Function with Zeros −2-2 and 33, and f(0)=12f(0) = 12

The function f(x)=−2(x+2)(x−3)f(x) = -2(x + 2)(x - 3), where the leading coefficient a=−2a = -2 satisfies f(0)=a(2)(−3)=12f(0) = a(2)(-3) = 12.

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Linear Factor for Zero −4-4

The factor (x+4)(x + 4), which must divide any polynomial that has a zero at x=−4x = -4.

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Quadratic Model for Points (−1,6)(-1, 6), (0,3)(0, 3), and (1,2)(1, 2)

The function f(x)=x2−2x+3f(x) = x^2 - 2x + 3, determined by solving for coefficients with c=3c = 3.

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Constant Term cc in f(x)=ax2+bx+cf(x) = ax^2 + bx + c

The yy-intercept of the quadratic function, which equals −4-4 when the graph contains the point (0,−4)(0, -4).

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Parabola with Vertex (1,−2)(1, -2) Passing Through (3,6)(3, 6)

The vertex form equation f(x)=2(x−1)2−2f(x) = 2(x - 1)^2 - 2, where a=2a = 2 is found using 6=a(3−1)2−26 = a(3 - 1)^2 - 2.

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Parabola Crossing xx-axis at −3-3 and 22 Passing Through (0,−12)(0, -12)

The equation f(x)=2(x+3)(x−2)f(x) = 2(x + 3)(x - 2), where a=2a = 2 satisfies −12=a(3)(−2)-12 = a(3)(-2).

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Vertex Form of a Quadratic Function

The form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k, which is most convenient to use when a graph provides its vertex (h,k)(h, k) and one other point.

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Area Function for Rectangular Garden with 80 m80\,\text{m} Fencing

The function A(x)=x(40−x)A(x) = x(40 - x), where xx is the length of one side in meters and the semi-perimeter is 40 m40\,\text{m}.

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Dimensions for Maximum Area with Perimeter 60 m60\,\text{m}

15 m by 15 m15\,\text{m}\text{ by }15\,\text{m}, which defines a square that maximizes rectangular area for a fixed perimeter.

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Greatest Possible Area for A(x)=−x2+36xA(x) = -x^2 + 36x

324 square units324\,\text{square units}, achieved at the vertex where x=18x = 18.

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Time of Maximum Height for h(t)=−5t2+40t+2h(t) = -5t^2 + 40t + 2

t=4 st = 4\,\text{s}, calculated from the vertex formula t=−b2a=−402(−5)t = -\frac{b}{2a} = -\frac{40}{2(-5)}.

38
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Maximum Height for h(t)=−4t2+24th(t) = -4t^2 + 24t

36 m36\,\text{m}, attained at the vertex where t=3 st = 3\,\text{s}.

39
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Profit Function P(x)P(x) for (50+x)(50 + x) Items at (100−2x)(100 - 2x) Pesos

The product function P(x)=(50+x)(100−2x)=−2x2+5000P(x) = (50 + x)(100 - 2x) = -2x^2 + 5000.

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Value of xx Maximizing Profit for P(x)=−2x2+80x+1000P(x) = -2x^2 + 80x + 1000

x=20x = 20, determined by finding the vertex coordinate x=−802(−2)x = -\frac{80}{2(-2)}.

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Vertex Meaning in Downward Profit Model at (30,4500)(30, 4500)

The coordinate 45004500 represents the maximum daily profit of P4,500\text{P}4,500, achieved when 3030 additional units are produced.

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Vertex Meaning in Downward Height Model at (2,18)(2, 18)

The physical interpretation that the object reaches its maximum height of 18 m18\,\text{m} after 2 s2\,\text{s}.

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Solution to x+7=5\sqrt{x + 7} = 5

x=18x = 18, obtained by squaring both sides to get x+7=25x + 7 = 25.

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Solution to 2x+1=82\sqrt{x + 1} = 8

x=15x = 15, obtained by isolating the radical to x+1=4\sqrt{x + 1} = 4 and squaring to get x+1=16x + 1 = 16.

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Solution to x+5=x−1\sqrt{x + 5} = x - 1

x=4x = 4, which is the only valid solution (since x=−1x = -1 produces an extraneous root).

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Equation x+4=−3\sqrt{x + 4} = -3

An equation that has no real solution because the principal square root of a real expression cannot be negative.

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General Form of (x−2)2+(y+1)2=16(x - 2)^2 + (y + 1)^2 = 16

The expanded circle equation x2+y2−4x+2y−11=0x^2 + y^2 - 4x + 2y - 11 = 0.

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Constant FF in General Form of (x+3)2+(y−4)2=25(x + 3)^2 + (y - 4)^2 = 25

F=0F = 0, since expanding gives x2+y2+6x−8y+(9+16−25)=0x^2 + y^2 + 6x - 8y + (9 + 16 - 25) = 0.

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Center-Radius Form of x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0

(x−3)2+(y+2)2=25(x - 3)^2 + (y + 2)^2 = 25, representing a circle centered at (3,−2)(3, -2) with a radius of 55.

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Center-Radius Form of x2+y2+2x−8y+8=0x^2 + y^2 + 2x - 8y + 8 = 0

(x+1)2+(y−4)2=9(x + 1)^2 + (y - 4)^2 = 9, representing a circle centered at (−1,4)(-1, 4) with a radius of 33.