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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.
Monic Quadratic Equation with Zeros 2 and −5
The quadratic equation f(x)=x2+3x−10, formed by expanding (x−2)(x+5) where the leading coefficient is 1.
Quadratic Equation with Zeros −2 and 3, and f(0)=12
The function f(x)=−2(x+2)(x−3), derived using factored form f(x)=a(x−r1)(x−r2) and solving for a=−2.
Linear Factor for Zeros −4 and 1
An expression such as (x+4) or (x−1) that must divide the quadratic equation with roots −4 and 1.
Quadratic Function Passing Through Points (−1,6), (0,3), and (1,2)
The function f(x)=x2−2x+3, which satisfies all three coordinate points.
y-Intercept Constant (c)
The constant term c in f(x)=ax2+bx+c, which equals −4 when the table contains (0,−4).
Quadratic Equation with Vertex (1,−2) Passing Through (3,6)
The quadratic function f(x)=2(x−1)2−2, determined using vertex form.
Parabola with Zeros −3 and 2 Passing Through (0,−12)
The quadratic equation f(x)=2(x+3)(x−2), derived using factored form with a=2.
Vertex Form of a Quadratic Function
The form f(x)=a(x−h)2+k, which is most convenient when a graph provides its vertex (h,k) and one other point.
Area Function of a Garden Enclosed by 80m of Fencing
The quadratic function A(x)=x(40−x), where x is the length of one side and (40−x) is the adjacent side.
Dimensions for Maximum Area with Perimeter 60m
15m by 15m, forming a square to maximize the rectangular area.
Greatest Possible Area for A(x)=−x2+36x
324square units, which occurs at the vertex where x=18.
Time of Maximum Height for h(t)=−5t2+40t+2
4s, calculated by evaluating the vertex time t=−2ab=−2(−5)40.
Maximum Height for h(t)=−4t2+24t
36m, reached at time t=3s.
Profit Model P(x) for (50+x) Items at (100−2x) Pesos
The quadratic model P(x)=−2x2+5000, calculated by expanding (50+x)(100−2x).
Optimal Production Level for P(x)=−2x2+80x+1000
x=20, the production value that maximizes profit at the vertex x=−2(−2)80.
Vertex y-Coordinate in Profit Model (30,4500)
The maximum daily profit of P4,500, corresponding to producing 30 additional units.
Vertex Coordinates in Projectile Height Model (2,18)
The point indicating that the object reaches its maximum height of 18m after 2s.
Solution to x+7=5
x=18, obtained by squaring both sides to get x+7=25.
Solution to 2x+1=8
x=15, determined by isolating the radical to x+1=4 and solving x+1=16.
Solution to x+5=x−1
x=4, which satisfies the equation, whereas x=−1 is extraneous.
Solvability of x+4=−3
It has no real solution because a principal square root cannot equal a negative number.
General Form of (x−2)2+(y+1)2=16
x2+y2−4x+2y−11=0, found by expanding and collecting all terms on one side.
Constant F in the General Form of (x+3)2+(y−4)2=25
0, calculated from 32+(−4)2−25=9+16−25.
Center-Radius Form of x2+y2−6x+4y−12=0
(x−3)2+(y+2)2=25, derived by completing the squares for x and y.
Center-Radius Form of x2+y2+2x−8y+8=0
(x+1)2+(y−4)2=9, determined by completing the squares for x and y.
Monic Quadratic with Zeros 2 and −5
The quadratic function f(x)=x2+3x−10, formed by expanding (x−2)(x+5).
Quadratic Function with Zeros −2 and 3, and f(0)=12
The function f(x)=−2(x+2)(x−3), where the leading coefficient a=−2 satisfies f(0)=a(2)(−3)=12.
Linear Factor for Zero −4
The factor (x+4), which must divide any polynomial that has a zero at x=−4.
Quadratic Model for Points (−1,6), (0,3), and (1,2)
The function f(x)=x2−2x+3, determined by solving for coefficients with c=3.
Constant Term c in f(x)=ax2+bx+c
The y-intercept of the quadratic function, which equals −4 when the graph contains the point (0,−4).
Parabola with Vertex (1,−2) Passing Through (3,6)
The vertex form equation f(x)=2(x−1)2−2, where a=2 is found using 6=a(3−1)2−2.
Parabola Crossing x-axis at −3 and 2 Passing Through (0,−12)
The equation f(x)=2(x+3)(x−2), where a=2 satisfies −12=a(3)(−2).
Vertex Form of a Quadratic Function
The form f(x)=a(x−h)2+k, which is most convenient to use when a graph provides its vertex (h,k) and one other point.
Area Function for Rectangular Garden with 80m Fencing
The function A(x)=x(40−x), where x is the length of one side in meters and the semi-perimeter is 40m.
Dimensions for Maximum Area with Perimeter 60m
15m by 15m, which defines a square that maximizes rectangular area for a fixed perimeter.
Greatest Possible Area for A(x)=−x2+36x
324square units, achieved at the vertex where x=18.
Time of Maximum Height for h(t)=−5t2+40t+2
t=4s, calculated from the vertex formula t=−2ab=−2(−5)40.
Maximum Height for h(t)=−4t2+24t
36m, attained at the vertex where t=3s.
Profit Function P(x) for (50+x) Items at (100−2x) Pesos
The product function P(x)=(50+x)(100−2x)=−2x2+5000.
Value of x Maximizing Profit for P(x)=−2x2+80x+1000
x=20, determined by finding the vertex coordinate x=−2(−2)80.
Vertex Meaning in Downward Profit Model at (30,4500)
The coordinate 4500 represents the maximum daily profit of P4,500, achieved when 30 additional units are produced.
Vertex Meaning in Downward Height Model at (2,18)
The physical interpretation that the object reaches its maximum height of 18m after 2s.
Solution to x+7=5
x=18, obtained by squaring both sides to get x+7=25.
Solution to 2x+1=8
x=15, obtained by isolating the radical to x+1=4 and squaring to get x+1=16.
Solution to x+5=x−1
x=4, which is the only valid solution (since x=−1 produces an extraneous root).
Equation x+4=−3
An equation that has no real solution because the principal square root of a real expression cannot be negative.
General Form of (x−2)2+(y+1)2=16
The expanded circle equation x2+y2−4x+2y−11=0.
Constant F in General Form of (x+3)2+(y−4)2=25
F=0, since expanding gives x2+y2+6x−8y+(9+16−25)=0.
Center-Radius Form of x2+y2−6x+4y−12=0
(x−3)2+(y+2)2=25, representing a circle centered at (3,−2) with a radius of 5.
Center-Radius Form of x2+y2+2x−8y+8=0
(x+1)2+(y−4)2=9, representing a circle centered at (−1,4) with a radius of 3.