SAT Math Flashcards: Quadratic Equations

Study Quadratic Equations in SAT Math with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

SAT Math

Quadratic Equations

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QUESTION
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Find the vertex of y=x2+4x3y = -x^2 + 4x - 3.

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ANSWER

Vertex: (2,1)(2, 1). Use x=b2a=2x = -\frac{b}{2a} = 2, then y=(4)+83=1y = -(4) + 8 - 3 = 1.

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What this deck covers

This deck focuses on Quadratic Equations, giving you a quick way to review the definitions, rules, and examples that matter most for SAT Math.

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Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

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Flashcard 1: Find the vertex of y=x2+4x3y = -x^2 + 4x - 3.

Answer: Vertex: (2,1)(2, 1). Use x=b2a=2x = -\frac{b}{2a} = 2, then y=(4)+83=1y = -(4) + 8 - 3 = 1.

Flashcard 2: Factor x29x^2 - 9.

Answer: (x3)(x+3)(x - 3)(x + 3). Difference of squares: a2b2=(ab)(a+b)a^2 - b^2 = (a-b)(a+b).

Flashcard 3: What is the quadratic formula?

Answer: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Derived from completing the square on the general quadratic equation.

Flashcard 4: What is the minimum value of y=x26x+9y = x^2 - 6x + 9?

Answer: Minimum: 00. This is (x3)2(x-3)^2, so minimum occurs at vertex (3,0)(3,0).

Flashcard 5: How do you determine the number of real roots using the discriminant?

Answer: If b24ac>0b^2 - 4ac > 0, 2 real roots; =0=0, 1 real root; <0<0, no real roots. The discriminant determines whether roots are real or complex.

Flashcard 6: What is the standard form of a quadratic equation?

Answer: ax2+bx+c=0ax^2 + bx + c = 0. General form where a0a \neq 0 determines parabola shape.

Flashcard 7: What is the minimum value of y=x26x+9y = x^2 - 6x + 9?

Answer: Minimum: 00. This is (x3)2(x-3)^2, so minimum occurs at vertex (3,0)(3,0).

Flashcard 8: State the quadratic formula.

Answer: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Solves any quadratic by substituting coefficients aa, bb, and cc.

Flashcard 9: Write x2+8x+16x^2 + 8x + 16 as a square.

Answer: (x+4)2(x + 4)^2. Perfect square trinomial with a=xa = x, b=4b = 4.

Flashcard 10: Solve for xx: x2+2x8=0x^2 + 2x - 8 = 0.

Answer: x=2,x=4x = 2, x = -4. Factor as (x2)(x+4)=0(x-2)(x+4) = 0 or use the quadratic formula.

Flashcard 11: What is the discriminant of ax2+bx+c=0ax^2 + bx + c = 0?

Answer: b24acb^2 - 4ac. The expression under the square root in the quadratic formula.

Flashcard 12: Factor x29x^2 - 9.

Answer: (x3)(x+3)(x - 3)(x + 3). Difference of squares: a2b2=(ab)(a+b)a^2 - b^2 = (a-b)(a+b).

Flashcard 13: For y=x2+4x+4y = x^2 + 4x + 4, what is the vertex?

Answer: Vertex: (2,0)(-2, 0). This is (x+2)2(x+2)^2, so vertex is at x=2x = -2, y=0y = 0.

Flashcard 14: What is the product of the roots of ax2+bx+c=0ax^2 + bx + c = 0?

Answer: ca\frac{c}{a}. By Vieta's formulas relating coefficients to roots.

Flashcard 15: Determine the parabola direction for y=5x23x+2y = 5x^2 - 3x + 2.

Answer: Opens upwards. Positive coefficient of x2x^2 means parabola opens upward.

Flashcard 16: Convert y=x24x+4y = x^2 - 4x + 4 to vertex form.

Answer: y=(x2)2y = (x-2)^2. This is (x2)2(x-2)^2 expanded, so vertex form shows vertex (2,0)(2,0).

Flashcard 17: For y=x2+4x+4y = x^2 + 4x + 4, what is the vertex?

Answer: Vertex: (2,0)(-2, 0). This is (x+2)2(x+2)^2, so vertex is at x=2x = -2, y=0y = 0.

Flashcard 18: How do you determine the number of real roots using the discriminant?

Answer: If b24ac>0b^2 - 4ac > 0, 2 real roots; =0=0, 1 real root; <0<0, no real roots. The discriminant determines whether roots are real or complex.

Flashcard 19: Identify the axis of symmetry for y=ax2+bx+cy = ax^2 + bx + c.

Answer: x=b2ax = \frac{-b}{2a}. Derived from completing the square or using calculus.

Flashcard 20: What is the quadratic formula?

Answer: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Derived from completing the square on the general quadratic equation.

Flashcard 21: What type of parabola does y=x2y = -x^2 represent?

Answer: A downward-opening parabola. Negative coefficient of x2x^2 makes the parabola open downward.

Flashcard 22: Simplify: (x+2)2(x + 2)^2.

Answer: x2+4x+4x^2 + 4x + 4. Apply the perfect square formula (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2.

Flashcard 23: Convert y=2(x3)2+4y = 2(x-3)^2 + 4 to standard form.

Answer: y=2x212x+22y = 2x^2 - 12x + 22. Expand (x3)2=x26x+9(x-3)^2 = x^2 - 6x + 9, then distribute and combine.

Flashcard 24: Convert y=x2+6x+8y = x^2 + 6x + 8 to vertex form.

Answer: y=(x+3)21y = (x+3)^2 - 1. Complete the square: (x+3)2=x2+6x+9(x+3)^2 = x^2 + 6x + 9, so subtract 1.

Flashcard 25: What is the vertex form of a quadratic equation?

Answer: y=a(xh)2+ky = a(x-h)^2 + k. Shows vertex (h,k)(h,k) and vertical shift directly.

Flashcard 26: Identify the axis of symmetry for y=ax2+bx+cy = ax^2 + bx + c.

Answer: x=b2ax = \frac{-b}{2a}. Derived from completing the square or using calculus.

Flashcard 27: What is the standard form of a quadratic equation?

Answer: ax2+bx+c=0ax^2 + bx + c = 0. General form where a0a \neq 0 determines parabola shape.

Flashcard 28: What is the standard form of a quadratic equation?

Answer: ax2+bx+c=0ax^2 + bx + c = 0. The general form where a0a \neq 0 defines a parabola.

Flashcard 29: What is the vertex of y=2(x1)2+3y = 2(x-1)^2 + 3?

Answer: Vertex: (1,3)(1, 3). In vertex form y=a(xh)2+ky = a(x-h)^2 + k, vertex is (h,k)(h,k).

Flashcard 30: State the zero-product property.

Answer: If ab=0ab = 0, then a=0a = 0 or b=0b = 0. If a product equals zero, at least one factor must be zero.

Flashcard 31: State the condition for a perfect square trinomial.

Answer: b2=4acb^2 = 4ac. When discriminant equals zero, the quadratic is a perfect square.

Flashcard 32: Find the x-intercepts of y=x24y = x^2 - 4.

Answer: x=2,x=2x = -2, x = 2. Set y=0y = 0: x24=0x^2 - 4 = 0, so x2=4x^2 = 4.

Flashcard 33: State the zero-product property.

Answer: If ab=0ab = 0, then a=0a = 0 or b=0b = 0. If a product equals zero, at least one factor must be zero.

Flashcard 34: Convert y=x24x+4y = x^2 - 4x + 4 to vertex form.

Answer: y=(x2)2y = (x-2)^2. This is (x2)2(x-2)^2 expanded, so vertex form shows vertex (2,0)(2,0).

Flashcard 35: What is the discriminant of ax2+bx+c=0ax^2 + bx + c = 0?

Answer: b24acb^2 - 4ac. The expression under the square root in the quadratic formula.

Flashcard 36: Write x2+8x+16x^2 + 8x + 16 as a square.

Answer: (x+4)2(x + 4)^2. Perfect square trinomial with a=xa = x, b=4b = 4.

Flashcard 37: Determine the parabola direction for y=5x23x+2y = 5x^2 - 3x + 2.

Answer: Opens upwards. Positive coefficient of x2x^2 means parabola opens upward.

Flashcard 38: Find the roots of x25x+6=0x^2 - 5x + 6 = 0.

Answer: x=2,x=3x = 2, x = 3. Factor as (x2)(x3)=0(x-2)(x-3) = 0 to find roots.

Flashcard 39: What is the axis of symmetry for y=ax2+bx+cy = ax^2 + bx + c?

Answer: x=b2ax = \frac{-b}{2a}. The xx-coordinate of the vertex, found by calculus or completing the square.

Flashcard 40: Identify the yy-intercept of y=2x2+3x+1y = 2x^2 + 3x + 1.

Answer: Intercept: (0,1)(0, 1). Set x=0x = 0 to find where the parabola crosses the yy-axis.

Flashcard 41: Find the discriminant for 2x24x+2=02x^2 - 4x + 2 = 0.

Answer: Discriminant: 00. Calculate b24ac=1616=0b^2 - 4ac = 16 - 16 = 0.

Flashcard 42: What does a positive discriminant indicate about roots?

Answer: Two distinct real roots. When b24ac>0b^2 - 4ac > 0, parabola crosses x-axis twice.

Flashcard 43: Solve 3x212=03x^2 - 12 = 0.

Answer: x=2,x=2x = 2, x = -2. Divide by 3: x24=0x^2 - 4 = 0, so x=±2x = \pm 2.

Flashcard 44: Convert y=x2+6x+8y = x^2 + 6x + 8 to vertex form.

Answer: y=(x+3)21y = (x+3)^2 - 1. Complete the square: (x+3)2=x2+6x+9(x+3)^2 = x^2 + 6x + 9, so subtract 1.

Flashcard 45: Find the axis of symmetry for y=3x2+6x+1y = 3x^2 + 6x + 1.

Answer: x=1x = -1. Use x=b2a=62(3)=1x = -\frac{b}{2a} = -\frac{6}{2(3)} = -1.

Flashcard 46: What is the effect of increasing aa in y=ax2y = ax^2 on the parabola?

Answer: Narrower parabola. Larger a|a| values compress the parabola horizontally.

Flashcard 47: What is the discriminant in the quadratic formula?

Answer: b24acb^2 - 4ac. Expression under the square root that determines root types.

Flashcard 48: What is the axis of symmetry for y=ax2+bx+cy = ax^2 + bx + c?

Answer: x=b2ax = \frac{-b}{2a}. The xx-coordinate of the vertex, found by calculus or completing the square.

Flashcard 49: What is the vertex of y=2(x1)2+3y = 2(x-1)^2 + 3?

Answer: Vertex: (1,3)(1, 3). In vertex form y=a(xh)2+ky = a(x-h)^2 + k, vertex is (h,k)(h,k).

Flashcard 50: What is the vertex form of a quadratic equation?

Answer: y=a(xh)2+ky = a(x-h)^2 + k. Shows vertex (h,k)(h,k) and vertical shift directly.

Flashcard 51: Which term in ax2+bx+cax^2 + bx + c affects the parabola's direction?

Answer: The coefficient aa. Positive aa opens up, negative aa opens down.

Flashcard 52: What is the standard form of a quadratic equation?

Answer: ax2+bx+c=0ax^2 + bx + c = 0. The general form where a0a \neq 0 defines a parabola.

Flashcard 53: What is the product of the roots of ax2+bx+c=0ax^2 + bx + c = 0?

Answer: ca\frac{c}{a}. By Vieta's formulas relating coefficients to roots.

Flashcard 54: Identify the yy-intercept of y=2x2+3x+1y = 2x^2 + 3x + 1.

Answer: Intercept: (0,1)(0, 1). Set x=0x = 0 to find where the parabola crosses the yy-axis.

Flashcard 55: State the quadratic formula.

Answer: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Solves any quadratic by substituting coefficients aa, bb, and cc.

Flashcard 56: Find the vertex of y=x24x+3y = x^2 - 4x + 3.

Answer: Vertex: (2,1)(2, -1). Use x=42=2x = \frac{4}{2} = 2, then y=48+3=1y = 4 - 8 + 3 = -1.

Flashcard 57: Find the discriminant for 2x24x+2=02x^2 - 4x + 2 = 0.

Answer: Discriminant: 00. Calculate b24ac=1616=0b^2 - 4ac = 16 - 16 = 0.

Flashcard 58: Which method can solve any quadratic equation?

Answer: The quadratic formula. Works for any quadratic, unlike factoring or square roots.

Flashcard 59: Find the vertex of y=x24x+3y = x^2 - 4x + 3.

Answer: Vertex: (2,1)(2, -1). Use x=42=2x = \frac{4}{2} = 2, then y=48+3=1y = 4 - 8 + 3 = -1.

Flashcard 60: What is the discriminant in the quadratic formula?

Answer: b24acb^2 - 4ac. Expression under the square root that determines root types.

Flashcard 61: Which term in ax2+bx+cax^2 + bx + c affects the parabola's direction?

Answer: The coefficient aa. Positive aa opens up, negative aa opens down.

Flashcard 62: Simplify: (x+2)2(x + 2)^2.

Answer: x2+4x+4x^2 + 4x + 4. Apply the perfect square formula (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2.

Flashcard 63: Identify the vertex form of a quadratic equation.

Answer: y=a(xh)2+ky = a(x-h)^2 + k. Shows vertex (h,k)(h,k) and vertical stretch/compression factor aa.

Flashcard 64: Which method can solve any quadratic equation?

Answer: The quadratic formula. Works for any quadratic, unlike factoring or square roots.

Flashcard 65: Find the x-intercepts of y=x24y = x^2 - 4.

Answer: x=2,x=2x = -2, x = 2. Set y=0y = 0: x24=0x^2 - 4 = 0, so x2=4x^2 = 4.

Flashcard 66: Identify the vertex form of a quadratic equation.

Answer: y=a(xh)2+ky = a(x-h)^2 + k. Shows vertex (h,k)(h,k) and vertical stretch/compression factor aa.

Flashcard 67: Find the roots of x25x+6=0x^2 - 5x + 6 = 0.

Answer: x=2,x=3x = 2, x = 3. Factor as (x2)(x3)=0(x-2)(x-3) = 0 or use the quadratic formula.

Flashcard 68: State the condition for a perfect square trinomial.

Answer: b2=4acb^2 = 4ac. When discriminant equals zero, the quadratic is a perfect square.

Flashcard 69: What is the sum of the roots of ax2+bx+c=0ax^2 + bx + c = 0?

Answer: ba-\frac{b}{a}. By Vieta's formulas relating coefficients to roots.

Flashcard 70: Find the axis of symmetry for y=3x2+6x+1y = 3x^2 + 6x + 1.

Answer: x=1x = -1. Use x=b2a=62(3)=1x = -\frac{b}{2a} = -\frac{6}{2(3)} = -1.

Flashcard 71: Convert y=2(x3)2+4y = 2(x-3)^2 + 4 to standard form.

Answer: y=2x212x+22y = 2x^2 - 12x + 22. Expand (x3)2=x26x+9(x-3)^2 = x^2 - 6x + 9, then distribute and combine.

Flashcard 72: What does a positive discriminant indicate about roots?

Answer: Two distinct real roots. When b24ac>0b^2 - 4ac > 0, parabola crosses x-axis twice.

Flashcard 73: Solve 3x212=03x^2 - 12 = 0.

Answer: x=2,x=2x = 2, x = -2. Divide by 3: x24=0x^2 - 4 = 0, so x=±2x = \pm 2.

Flashcard 74: Find the roots of x25x+6=0x^2 - 5x + 6 = 0.

Answer: x=2,x=3x = 2, x = 3. Factor as (x2)(x3)=0(x-2)(x-3) = 0 to find roots.