Study Solving Nonlinear Functions in SAT Math with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: Identify the axis of symmetry for y = x 2 + 6 x + 5 y = x^2 + 6x + 5 y = x 2 + 6 x + 5 . Answer: x = − 3 x = -3 x = − 3 . Use formula x = − b 2 a = − 6 2 ( 1 ) = − 3 x = -\frac{b}{2a} = -\frac{6}{2(1)} = -3 x = − 2 a b = − 2 ( 1 ) 6 = − 3 .
Flashcard 2: What is the effect of a < 0 a < 0 a < 0 in y = a ( x − h ) 2 + k y = a(x-h)^2 + k y = a ( x − h ) 2 + k ? Answer: Parabola opens downward. Negative coefficient makes parabola open downward.
Flashcard 3: Identify the roots of x 2 − 6 x + 9 = 0 x^2 - 6x + 9 = 0 x 2 − 6 x + 9 = 0 . Answer: x = 3 x = 3 x = 3 . Perfect square: ( x − 3 ) 2 = 0 (x-3)^2 = 0 ( x − 3 ) 2 = 0 has repeated root.
Flashcard 4: What is the effect of a > 0 a > 0 a > 0 in y = a ( x − h ) 2 + k y = a(x-h)^2 + k y = a ( x − h ) 2 + k ? Answer: Parabola opens upward. Positive coefficient makes parabola open upward.
Flashcard 5: What does the discriminant b 2 − 4 a c b^2 - 4ac b 2 − 4 a c indicate about roots? Answer: Number and type of roots. Determines number of real roots and their nature.
Flashcard 6: Which form is y = a ( x − h ) 2 + k y = a(x-h)^2 + k y = a ( x − h ) 2 + k ? Answer: Vertex form. Form that shows vertex ( h , k ) (h,k) ( h , k ) and transformations.
Flashcard 7: Convert the quadratic y = 2 x 2 + 8 x + 6 y = 2x^2 + 8x + 6 y = 2 x 2 + 8 x + 6 to vertex form. Answer: y = 2 ( x + 2 ) 2 − 2 y = 2(x+2)^2 - 2 y = 2 ( x + 2 ) 2 − 2 . Complete the square: factor out 2, then add/subtract 4.
Flashcard 8: If y = x 2 + 2 x + 1 y = x^2 + 2x + 1 y = x 2 + 2 x + 1 , what is the axis of symmetry? Answer: x = − 1 x = -1 x = − 1 . Perfect square: ( x + 1 ) 2 (x+1)^2 ( x + 1 ) 2 , so axis is x = − 1 x = -1 x = − 1 .
Flashcard 9: Convert y = x 2 + 6 x + 9 y = x^2 + 6x + 9 y = x 2 + 6 x + 9 to vertex form. Answer: y = ( x + 3 ) 2 y = (x+3)^2 y = ( x + 3 ) 2 . Perfect square: ( x + 3 ) 2 = x 2 + 6 x + 9 (x+3)^2 = x^2 + 6x + 9 ( x + 3 ) 2 = x 2 + 6 x + 9 .
Flashcard 10: What is the solution to 4 x 2 − 4 x + 1 = 0 4x^2 - 4x + 1 = 0 4 x 2 − 4 x + 1 = 0 ? Answer: x = 1 2 x = \frac{1}{2} x = 2 1 . Perfect square: ( 2 x − 1 ) 2 = 0 (2x-1)^2 = 0 ( 2 x − 1 ) 2 = 0 gives x = 1 2 x = \frac{1}{2} x = 2 1 .
Flashcard 11: Find the roots of x 2 − 5 x + 6 = 0 x^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0 using factoring. Answer: x = 2 , x = 3 x = 2, x = 3 x = 2 , x = 3 . Factors as ( x − 2 ) ( x − 3 ) = 0 (x-2)(x-3) = 0 ( x − 2 ) ( x − 3 ) = 0 , so x = 2 x = 2 x = 2 or x = 3 x = 3 x = 3 .
Flashcard 12: State the relationship between roots and factors of a quadratic. Answer: Roots are solutions; factors are ( x − r 1 ) ( x − r 2 ) (x - r_1)(x - r_2) ( x − r 1 ) ( x − r 2 ) . If roots are r 1 , r 2 r_1, r_2 r 1 , r 2 , then factors are ( x − r 1 ) ( x − r 2 ) (x-r_1)(x-r_2) ( x − r 1 ) ( x − r 2 ) .
Flashcard 13: What is the solution for 4 x 2 = 16 4x^2 = 16 4 x 2 = 16 ? Answer: x = 2 x = 2 x = 2 or x = − 2 x = -2 x = − 2 . Divide by 4: x 2 = 4 x^2 = 4 x 2 = 4 , so x = ± 2 x = \pm 2 x = ± 2 .
Flashcard 14: State the quadratic formula. Answer: x = − b ± b 2 − 4 a c 2 a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x = 2 a − b ± b 2 − 4 a c . Standard formula for solving quadratic equations.
Flashcard 15: What is the discriminant in the quadratic formula? Answer: b 2 − 4 a c b^2 - 4ac b 2 − 4 a c . Expression under the square root that determines root types.
Flashcard 16: Convert y = x 2 + 4 x + 4 y = x^2 + 4x + 4 y = x 2 + 4 x + 4 to vertex form. Answer: y = ( x + 2 ) 2 y = (x+2)^2 y = ( x + 2 ) 2 . Complete the square: ( x + 2 ) 2 = x 2 + 4 x + 4 (x+2)^2 = x^2 + 4x + 4 ( x + 2 ) 2 = x 2 + 4 x + 4 .
Flashcard 17: What is the solution for 4 x 2 = 16 4x^2 = 16 4 x 2 = 16 ? Answer: x = 2 x = 2 x = 2 or x = − 2 x = -2 x = − 2 . Divide by 4: x 2 = 4 x^2 = 4 x 2 = 4 , so x = ± 2 x = \pm 2 x = ± 2 .
Flashcard 18: What is the standard form of a quadratic equation? Answer: a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 . General form with degree 2 polynomial set to zero.
Flashcard 19: Determine the nature of roots for x 2 + 2 x + 5 = 0 x^2 + 2x + 5 = 0 x 2 + 2 x + 5 = 0 . Answer: Complex. Discriminant 4 − 20 = − 16 < 0 4 - 20 = -16 < 0 4 − 20 = − 16 < 0 means complex roots.
Flashcard 20: What is the vertex of y = x 2 − 2 x + 1 y = x^2 - 2x + 1 y = x 2 − 2 x + 1 ? Answer: ( 1 , 0 ) (1, 0) ( 1 , 0 ) . Perfect square: ( x − 1 ) 2 (x-1)^2 ( x − 1 ) 2 has vertex at ( 1 , 0 ) (1,0) ( 1 , 0 ) .
Flashcard 21: Identify the discriminant of x 2 + 4 x + 4 = 0 x^2 + 4x + 4 = 0 x 2 + 4 x + 4 = 0 . Answer: 0 0 0 . Calculate b 2 − 4 a c = 16 − 16 = 0 b^2 - 4ac = 16 - 16 = 0 b 2 − 4 a c = 16 − 16 = 0 .
Flashcard 22: Convert y = x 2 + 4 x + 4 y = x^2 + 4x + 4 y = x 2 + 4 x + 4 to vertex form. Answer: y = ( x + 2 ) 2 y = (x+2)^2 y = ( x + 2 ) 2 . Complete the square: ( x + 2 ) 2 = x 2 + 4 x + 4 (x+2)^2 = x^2 + 4x + 4 ( x + 2 ) 2 = x 2 + 4 x + 4 .
Flashcard 23: Solve for x x x : x 2 + 9 = 0 x^2 + 9 = 0 x 2 + 9 = 0 . Answer: x = 3 i x = 3i x = 3 i or x = − 3 i x = -3i x = − 3 i . Take square root: x 2 = − 9 x^2 = -9 x 2 = − 9 , so x = ± 3 i x = \pm 3i x = ± 3 i .
Flashcard 24: Which form is y = a ( x − p ) ( x − q ) y = a(x-p)(x-q) y = a ( x − p ) ( x − q ) ? Answer: Factored form. Shows roots p p p and q q q directly as factors.
Flashcard 25: What is a nonlinear function? Answer: A function that is not a straight line. Graph is not a straight line (degree > 1).
Flashcard 26: Solve for x x x : x 2 + 6 x + 9 = 0 x^2 + 6x + 9 = 0 x 2 + 6 x + 9 = 0 using the square root method. Answer: x = − 3 x = -3 x = − 3 . Perfect square trinomial: ( x + 3 ) 2 = 0 (x+3)^2 = 0 ( x + 3 ) 2 = 0 , so x = − 3 x = -3 x = − 3 .
Flashcard 27: Find the discriminant of x 2 − 4 x + 4 = 0 x^2 - 4x + 4 = 0 x 2 − 4 x + 4 = 0 . Answer: 0 0 0 . Calculate b 2 − 4 a c = 16 − 16 = 0 b^2 - 4ac = 16 - 16 = 0 b 2 − 4 a c = 16 − 16 = 0 .
Flashcard 28: Identify the maximum or minimum value of y = ( x − 2 ) 2 + 5 y = (x-2)^2 + 5 y = ( x − 2 ) 2 + 5 . Answer: Minimum: 5 5 5 . Since a > 0 a > 0 a > 0 , parabola opens up with minimum at vertex.
Flashcard 29: What is the sum of the roots of a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 ? Answer: − b a -\frac{b}{a} − a b . Vieta's formula for sum of roots.
Flashcard 30: Identify the discriminant of x 2 + 4 x + 4 = 0 x^2 + 4x + 4 = 0 x 2 + 4 x + 4 = 0 . Answer: 0 0 0 . Calculate b 2 − 4 a c = 16 − 16 = 0 b^2 - 4ac = 16 - 16 = 0 b 2 − 4 a c = 16 − 16 = 0 .
Flashcard 31: Determine the nature of roots for x 2 + 2 x + 5 = 0 x^2 + 2x + 5 = 0 x 2 + 2 x + 5 = 0 . Answer: Complex. Discriminant 4 − 20 = − 16 < 0 4 - 20 = -16 < 0 4 − 20 = − 16 < 0 means complex roots.
Flashcard 32: Find the vertex of y = − x 2 + 6 x − 9 y = -x^2 + 6x - 9 y = − x 2 + 6 x − 9 . Answer: ( 3 , 0 ) (3, 0) ( 3 , 0 ) . Perfect square: − ( x − 3 ) 2 -(x-3)^2 − ( x − 3 ) 2 has vertex at ( 3 , 0 ) (3,0) ( 3 , 0 ) .
Flashcard 33: State the quadratic formula used to solve a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 . Answer: x = − b ± b 2 − 4 a c 2 a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x = 2 a − b ± b 2 − 4 a c . Derived by completing the square on the general quadratic form.
Flashcard 34: What is the solution to 4 x 2 − 4 x + 1 = 0 4x^2 - 4x + 1 = 0 4 x 2 − 4 x + 1 = 0 ? Answer: x = 1 2 x = \frac{1}{2} x = 2 1 . Perfect square: ( 2 x − 1 ) 2 = 0 (2x-1)^2 = 0 ( 2 x − 1 ) 2 = 0 gives x = 1 2 x = \frac{1}{2} x = 2 1 .
Flashcard 35: What is the vertex of y = − 2 ( x + 3 ) 2 + 4 y = -2(x+3)^2 + 4 y = − 2 ( x + 3 ) 2 + 4 ? Answer: ( − 3 , 4 ) (-3, 4) ( − 3 , 4 ) . Vertex form shows ( h , k ) = ( − 3 , 4 ) (h,k) = (-3,4) ( h , k ) = ( − 3 , 4 ) directly.
Flashcard 36: What is the standard form equation of a quadratic function? Answer: a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 . General form where a ≠ 0 a \neq 0 a = 0 defines any quadratic equation.
Flashcard 37: State the quadratic formula. Answer: x = − b ± b 2 − 4 a c 2 a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x = 2 a − b ± b 2 − 4 a c . Standard formula for solving quadratic equations.
Flashcard 38: Find the roots of x 2 − 5 x + 6 = 0 x^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0 using factoring. Answer: x = 2 , x = 3 x = 2, x = 3 x = 2 , x = 3 . Factors as ( x − 2 ) ( x − 3 ) = 0 (x-2)(x-3) = 0 ( x − 2 ) ( x − 3 ) = 0 , so x = 2 x = 2 x = 2 or x = 3 x = 3 x = 3 .
Flashcard 39: Identify the axis of symmetry for y = x 2 + 6 x + 5 y = x^2 + 6x + 5 y = x 2 + 6 x + 5 . Answer: x = − 3 x = -3 x = − 3 . Use formula x = − b 2 a = − 6 2 ( 1 ) = − 3 x = -\frac{b}{2a} = -\frac{6}{2(1)} = -3 x = − 2 a b = − 2 ( 1 ) 6 = − 3 .
Flashcard 40: State the relationship between roots and factors of a quadratic. Answer: Roots are solutions; factors are ( x − r 1 ) ( x − r 2 ) (x - r_1)(x - r_2) ( x − r 1 ) ( x − r 2 ) . If roots are r 1 , r 2 r_1, r_2 r 1 , r 2 , then factors are ( x − r 1 ) ( x − r 2 ) (x-r_1)(x-r_2) ( x − r 1 ) ( x − r 2 ) .
Flashcard 41: Find the roots of x 2 − 5 x + 6 = 0 x^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0 . Answer: x = 2 x = 2 x = 2 or x = 3 x = 3 x = 3 . Factor: ( x − 2 ) ( x − 3 ) = 0 (x-2)(x-3) = 0 ( x − 2 ) ( x − 3 ) = 0 gives roots.
Flashcard 42: Which form is y = a ( x − p ) ( x − q ) y = a(x-p)(x-q) y = a ( x − p ) ( x − q ) ? Answer: Factored form. Shows roots p p p and q q q directly as factors.
Flashcard 43: Solve for x x x : x 2 − 9 = 0 x^2 - 9 = 0 x 2 − 9 = 0 . Answer: x = 3 x = 3 x = 3 or x = − 3 x = -3 x = − 3 . Difference of squares: ( x − 3 ) ( x + 3 ) = 0 (x-3)(x+3) = 0 ( x − 3 ) ( x + 3 ) = 0 .
Flashcard 44: What are the roots of x 2 − 1 = 0 x^2 - 1 = 0 x 2 − 1 = 0 ? Answer: x = 1 x = 1 x = 1 or x = − 1 x = -1 x = − 1 . Difference of squares: ( x − 1 ) ( x + 1 ) = 0 (x-1)(x+1) = 0 ( x − 1 ) ( x + 1 ) = 0 .
Flashcard 45: Solve for x x x : x 2 − 4 = 0 x^2 - 4 = 0 x 2 − 4 = 0 . Answer: x = 2 x = 2 x = 2 or x = − 2 x = -2 x = − 2 . Factor as difference of squares: ( x − 2 ) ( x + 2 ) = 0 (x-2)(x+2) = 0 ( x − 2 ) ( x + 2 ) = 0 .
Flashcard 46: Solve for x x x : x 2 + 6 x + 9 = 0 x^2 + 6x + 9 = 0 x 2 + 6 x + 9 = 0 using the square root method. Answer: x = − 3 x = -3 x = − 3 . Perfect square trinomial: ( x + 3 ) 2 = 0 (x+3)^2 = 0 ( x + 3 ) 2 = 0 , so x = − 3 x = -3 x = − 3 .
Flashcard 47: Identify the roots of x 2 − 6 x + 9 = 0 x^2 - 6x + 9 = 0 x 2 − 6 x + 9 = 0 . Answer: x = 3 x = 3 x = 3 . Perfect square: ( x − 3 ) 2 = 0 (x-3)^2 = 0 ( x − 3 ) 2 = 0 has repeated root.
Flashcard 48: Identify the vertex form of a quadratic function. Answer: y = a ( x − h ) 2 + k y = a(x-h)^2 + k y = a ( x − h ) 2 + k . Shows vertex ( h , k ) (h,k) ( h , k ) and stretch factor a a a explicitly.
Flashcard 49: What is the product of the roots of a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 ? Answer: c a \frac{c}{a} a c . Vieta's formula for product of roots.
Flashcard 50: Convert the quadratic y = 2 x 2 + 8 x + 6 y = 2x^2 + 8x + 6 y = 2 x 2 + 8 x + 6 to vertex form. Answer: y = 2 ( x + 2 ) 2 − 2 y = 2(x+2)^2 - 2 y = 2 ( x + 2 ) 2 − 2 . Complete the square: factor out 2, then add/subtract 4.
Flashcard 51: Which method can solve x 2 + 4 x + 4 = 0 x^2 + 4x + 4 = 0 x 2 + 4 x + 4 = 0 besides factoring? Answer: Completing the square. Transforms equation to perfect square trinomial form.
Flashcard 52: Find the value of x x x in x 2 + 4 x + 4 = 0 x^2 + 4x + 4 = 0 x 2 + 4 x + 4 = 0 . Answer: x = − 2 x = -2 x = − 2 . Perfect square: ( x + 2 ) 2 = 0 (x+2)^2 = 0 ( x + 2 ) 2 = 0 gives x = − 2 x = -2 x = − 2 .
Flashcard 53: What does a positive discriminant indicate about a quadratic's roots? Answer: Two distinct real roots. When b 2 − 4 a c > 0 b^2 - 4ac > 0 b 2 − 4 a c > 0 , the parabola crosses the x-axis twice.
Flashcard 54: What is the vertex form of a quadratic function? Answer: y = a ( x − h ) 2 + k y = a(x-h)^2 + k y = a ( x − h ) 2 + k . Shows vertex ( h , k ) (h,k) ( h , k ) and vertical shifts/stretches.
Flashcard 55: What is the discriminant in the quadratic formula? Answer: b 2 − 4 a c b^2 - 4ac b 2 − 4 a c . Expression under the square root that determines root types.
Flashcard 56: What is the standard form of a quadratic equation? Answer: a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 . General form with degree 2 polynomial set to zero.
Flashcard 57: What is a nonlinear function? Answer: A function that is not a straight line. Graph is not a straight line (degree > 1).
Flashcard 58: For y = 3 ( x − 4 ) 2 − 7 y = 3(x-4)^2 - 7 y = 3 ( x − 4 ) 2 − 7 , what is the vertex? Answer: ( 4 , − 7 ) (4, -7) ( 4 , − 7 ) . Vertex form shows ( h , k ) = ( 4 , − 7 ) (h,k) = (4,-7) ( h , k ) = ( 4 , − 7 ) directly.
Flashcard 59: Convert y = x 2 + 6 x + 9 y = x^2 + 6x + 9 y = x 2 + 6 x + 9 to vertex form. Answer: y = ( x + 3 ) 2 y = (x+3)^2 y = ( x + 3 ) 2 . Perfect square: ( x + 3 ) 2 = x 2 + 6 x + 9 (x+3)^2 = x^2 + 6x + 9 ( x + 3 ) 2 = x 2 + 6 x + 9 .
Flashcard 60: Find the vertex of y = − x 2 + 6 x − 9 y = -x^2 + 6x - 9 y = − x 2 + 6 x − 9 . Answer: ( 3 , 0 ) (3, 0) ( 3 , 0 ) . Perfect square: − ( x − 3 ) 2 -(x-3)^2 − ( x − 3 ) 2 has vertex at ( 3 , 0 ) (3,0) ( 3 , 0 ) .
Flashcard 61: Solve for x x x : x 2 − 4 = 0 x^2 - 4 = 0 x 2 − 4 = 0 . Answer: x = 2 x = 2 x = 2 or x = − 2 x = -2 x = − 2 . Factor as difference of squares: ( x − 2 ) ( x + 2 ) = 0 (x-2)(x+2) = 0 ( x − 2 ) ( x + 2 ) = 0 .
Flashcard 62: What is the degree of a quadratic function? Answer:
Highest power of variable is 2.
Flashcard 63: What type of solutions does a quadratic have if the discriminant is negative? Answer: Complex solutions. Negative discriminant means no real roots.
Flashcard 64: What does a positive discriminant indicate about a quadratic's roots? Answer: Two distinct real roots. When b 2 − 4 a c > 0 b^2 - 4ac > 0 b 2 − 4 a c > 0 , the parabola crosses the x-axis twice.
Flashcard 65: What is the vertex of y = − 2 ( x + 3 ) 2 + 4 y = -2(x+3)^2 + 4 y = − 2 ( x + 3 ) 2 + 4 ? Answer: ( − 3 , 4 ) (-3, 4) ( − 3 , 4 ) . Vertex form shows ( h , k ) = ( − 3 , 4 ) (h,k) = (-3,4) ( h , k ) = ( − 3 , 4 ) directly.
Flashcard 66: What is the effect of a < 0 a < 0 a < 0 in y = a ( x − h ) 2 + k y = a(x-h)^2 + k y = a ( x − h ) 2 + k ? Answer: Parabola opens downward. Negative coefficient makes parabola open downward.
Flashcard 67: Find the vertex of y = 2 ( x − 3 ) 2 + 4 y = 2(x - 3)^2 + 4 y = 2 ( x − 3 ) 2 + 4 . Answer: Vertex: ( 3 , 4 ) (3, 4) ( 3 , 4 ) . In vertex form, ( h , k ) (h,k) ( h , k ) gives the vertex coordinates directly.
Flashcard 68: For y = ( x − 1 ) 2 + 3 y = (x-1)^2 + 3 y = ( x − 1 ) 2 + 3 , identify the vertex. Answer: ( 1 , 3 ) (1, 3) ( 1 , 3 ) . Vertex form shows ( h , k ) = ( 1 , 3 ) (h,k) = (1,3) ( h , k ) = ( 1 , 3 ) directly.
Flashcard 69: If y = x 2 + 2 x + 1 y = x^2 + 2x + 1 y = x 2 + 2 x + 1 , what is the axis of symmetry? Answer: x = − 1 x = -1 x = − 1 . Perfect square: ( x + 1 ) 2 (x+1)^2 ( x + 1 ) 2 , so axis is x = − 1 x = -1 x = − 1 .
Flashcard 70: Find the roots of x 2 − 5 x + 6 = 0 x^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0 . Answer: x = 2 x = 2 x = 2 or x = 3 x = 3 x = 3 . Factor: ( x − 2 ) ( x − 3 ) = 0 (x-2)(x-3) = 0 ( x − 2 ) ( x − 3 ) = 0 gives roots.
Flashcard 71: Convert y = x 2 − 4 x + 6 y = x^2 - 4x + 6 y = x 2 − 4 x + 6 to vertex form. Answer: y = ( x − 2 ) 2 + 2 y = (x-2)^2 + 2 y = ( x − 2 ) 2 + 2 . Complete the square: ( x − 2 ) 2 = x 2 − 4 x + 4 (x-2)^2 = x^2 - 4x + 4 ( x − 2 ) 2 = x 2 − 4 x + 4 .
Flashcard 72: What are the roots of x 2 − 1 = 0 x^2 - 1 = 0 x 2 − 1 = 0 ? Answer: x = 1 x = 1 x = 1 or x = − 1 x = -1 x = − 1 . Difference of squares: ( x − 1 ) ( x + 1 ) = 0 (x-1)(x+1) = 0 ( x − 1 ) ( x + 1 ) = 0 .
Flashcard 73: Find the vertex of y = 2 ( x − 3 ) 2 + 4 y = 2(x - 3)^2 + 4 y = 2 ( x − 3 ) 2 + 4 . Answer: Vertex: ( 3 , 4 ) (3, 4) ( 3 , 4 ) . In vertex form, ( h , k ) (h,k) ( h , k ) gives the vertex coordinates directly.
Flashcard 74: What type of solutions does a quadratic have if the discriminant is negative? Answer: Complex solutions. Negative discriminant means no real roots.
Flashcard 75: What is the product of the roots of a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 ? Answer: c a \frac{c}{a} a c . Vieta's formula for product of roots.
Flashcard 76: Identify the maximum or minimum value of y = ( x − 2 ) 2 + 5 y = (x-2)^2 + 5 y = ( x − 2 ) 2 + 5 . Answer: Minimum: 5 5 5 . Since a > 0 a > 0 a > 0 , parabola opens up with minimum at vertex.
Flashcard 77: Which form is y = a ( x − h ) 2 + k y = a(x-h)^2 + k y = a ( x − h ) 2 + k ? Answer: Vertex form. Form that shows vertex ( h , k ) (h,k) ( h , k ) and transformations.
Flashcard 78: State the quadratic formula used to solve a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 . Answer: x = − b ± b 2 − 4 a c 2 a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x = 2 a − b ± b 2 − 4 a c . Derived by completing the square on the general quadratic form.
Flashcard 79: Find the value of x x x in x 2 + 4 x + 4 = 0 x^2 + 4x + 4 = 0 x 2 + 4 x + 4 = 0 . Answer: x = − 2 x = -2 x = − 2 . Perfect square: ( x + 2 ) 2 = 0 (x+2)^2 = 0 ( x + 2 ) 2 = 0 gives x = − 2 x = -2 x = − 2 .
Flashcard 80: Find the axis of symmetry for y = 2 x 2 + 4 x + 1 y = 2x^2 + 4x + 1 y = 2 x 2 + 4 x + 1 . Answer: x = − 1 x = -1 x = − 1 . Use x = − b 2 a = − 4 2 ( 2 ) = − 1 x = -\frac{b}{2a} = -\frac{4}{2(2)} = -1 x = − 2 a b = − 2 ( 2 ) 4 = − 1 .
Flashcard 81: Solve for x x x : x 2 + 9 = 0 x^2 + 9 = 0 x 2 + 9 = 0 . Answer: x = 3 i x = 3i x = 3 i or x = − 3 i x = -3i x = − 3 i . Take square root: x 2 = − 9 x^2 = -9 x 2 = − 9 , so x = ± 3 i x = \pm 3i x = ± 3 i .
Flashcard 82: For y = ( x − 1 ) 2 + 3 y = (x-1)^2 + 3 y = ( x − 1 ) 2 + 3 , identify the vertex. Answer: ( 1 , 3 ) (1, 3) ( 1 , 3 ) . Vertex form shows ( h , k ) = ( 1 , 3 ) (h,k) = (1,3) ( h , k ) = ( 1 , 3 ) directly.
Flashcard 83: For y = 3 ( x − 4 ) 2 − 7 y = 3(x-4)^2 - 7 y = 3 ( x − 4 ) 2 − 7 , what is the vertex? Answer: ( 4 , − 7 ) (4, -7) ( 4 , − 7 ) . Vertex form shows ( h , k ) = ( 4 , − 7 ) (h,k) = (4,-7) ( h , k ) = ( 4 , − 7 ) directly.
Flashcard 84: What is the vertex form of a quadratic function? Answer: y = a ( x − h ) 2 + k y = a(x-h)^2 + k y = a ( x − h ) 2 + k . Shows vertex ( h , k ) (h,k) ( h , k ) and vertical shifts/stretches.
Flashcard 85: What is the degree of a quadratic function? Answer:
Highest power of variable is 2.
Flashcard 86: What does the discriminant b 2 − 4 a c b^2 - 4ac b 2 − 4 a c indicate about roots? Answer: Number and type of roots. Determines number of real roots and their nature.
Flashcard 87: What is the effect of a > 0 a > 0 a > 0 in y = a ( x − h ) 2 + k y = a(x-h)^2 + k y = a ( x − h ) 2 + k ? Answer: Parabola opens upward. Positive coefficient makes parabola open upward.
Flashcard 88: What is the vertex of y = x 2 − 2 x + 1 y = x^2 - 2x + 1 y = x 2 − 2 x + 1 ? Answer: ( 1 , 0 ) (1, 0) ( 1 , 0 ) . Perfect square: ( x − 1 ) 2 (x-1)^2 ( x − 1 ) 2 has vertex at ( 1 , 0 ) (1,0) ( 1 , 0 ) .
Flashcard 89: Identify the vertex form of a quadratic function. Answer: y = a ( x − h ) 2 + k y = a(x-h)^2 + k y = a ( x − h ) 2 + k . Shows vertex ( h , k ) (h,k) ( h , k ) and stretch factor a a a explicitly.
Flashcard 90: Find the discriminant of x 2 − 4 x + 4 = 0 x^2 - 4x + 4 = 0 x 2 − 4 x + 4 = 0 . Answer: 0 0 0 . Calculate b 2 − 4 a c = 16 − 16 = 0 b^2 - 4ac = 16 - 16 = 0 b 2 − 4 a c = 16 − 16 = 0 .
Flashcard 91: For y = x 2 − 4 x + 4 y = x^2 - 4x + 4 y = x 2 − 4 x + 4 , determine the axis of symmetry. Answer: x = 2 x = 2 x = 2 . Axis of symmetry is x = − b 2 a = − − 4 2 ( 1 ) = 2 x = -\frac{b}{2a} = -\frac{-4}{2(1)} = 2 x = − 2 a b = − 2 ( 1 ) − 4 = 2 .
Flashcard 92: What is the sum of the roots of a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 ? Answer: − b a -\frac{b}{a} − a b . Vieta's formula for sum of roots.