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Graphing Quadratics Using the Discriminant>
Graphing Quadratics Using the DiscriminantGraphing Quadratics Using the Discriminant
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Question 1 of 7
1. Question
Which of the following graphs have a discriminant that is equal to 00?
(Δ=0Δ=0)- 1.
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Nature of the Roots Discriminant (ΔΔ) Two real roots ΔΔ>>00 One real root Δ=0Δ=0 No real roots ΔΔ<<00 Discriminant Formula
Δ=b2−4acΔ=b2−4acFor each graph, check how many times the graph has passed through the xx axisThis graph touched the xx axis once, which means it has one rootHence, Δ=0Δ=0This graph touched the xx axis twice, which means it has two rootsHence, ΔΔ>>00This graph touched the xx axis once, which means it has one rootHence, Δ=0Δ=0This graph touched the xx axis twice, which means it has two rootsHence, ΔΔ>>00 -
Question 2 of 7
2. Question
Which function does not intersect with the xx axis?-
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Nature of the Roots Discriminant (ΔΔ) Two real roots ΔΔ>>00 One real root Δ=0Δ=0 No real roots ΔΔ<<00 Discriminant Formula
Δ=b2−4acΔ=b2−4acCompute for the discriminant of each functionx2-4x+3=0x2−4x+3=0a=1a=1 b=-4b=−4 c=3c=3ΔΔ == b2−4acb2−4ac Discriminant Formula == (−4)2−4(1)(3)(−4)2−4(1)(3) Substitute values == 16-1216−12 == 44 Since ΔΔ>>00, this function has two real roots and intersects the xx axis twicex2-4x+4=0x2−4x+4=0a=1a=1 b=-4b=−4 c=4c=4ΔΔ == b2−4acb2−4ac Discriminant Formula == (−4)2−4(1)(4)(−4)2−4(1)(4) Substitute values == 16-1616−16 == 00 Since Δ=0Δ=0, this function has one real root and intersects the xx axis oncex2-4x+5=0x2−4x+5=0a=1a=1 b=-4b=−4 c=5c=5ΔΔ == b2−4acb2−4ac Discriminant Formula == (−4)2−4(1)(5)(−4)2−4(1)(5) Substitute values == 16-2016−20 == -4−4 Since ΔΔ<<00, this function has no real roots and does not intersect the xx axisx2-4x+5=0x2−4x+5=0 -
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Question 3 of 7
3. Question
Graph using the discriminanty=2x2-2x-5y=2x2−2x−5-
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Chapters- Chapters
Quadratic Formula
x=−b±√Δ2ax=−b±√Δ2aDiscriminant Formula
Δ=b2−4acΔ=b2−4acFirst, compute for the discriminanty=2x2-2x-5y=2x2−2x−5a=2a=2 b=-2b=−2 c=-5c=−5ΔΔ == b2−4acb2−4ac Discriminant Formula == −22−4(2)(−5)−22−4(2)(−5) Substitute values == 4+404+40 == 4444 Next, substitute the discriminant to the Quadratic Formula to find the xx interceptsxx == −b±√Δ2a−b±√Δ2a Quadratic Formula == −(−2)±√442(2)−(−2)±√442(2) Substitute values == 2±2√1142±2√114 == 1±√1121±√112 Write each root individuallyx1x1 == 1+√1121+√112 == 2.1582.158 x1x1 == 1-√1121−√112 == −1.158−1.158 Mark these 22 points on the xx axisFind the yy intercept by substituting x=0x=0yy == 2x2-2x-52x2−2x−5 == 2(0)2-2(0)-52(0)2−2(0)−5 Substitute x=0x=0 == 0-0-50−0−5 == -5−5 Mark this on the yy axisFinally, form the parabola by connecting the points -
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Question 4 of 7
4. Question
Graph using the discriminanty=3x2-4x-4y=3x2−4x−4-
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Quadratic Formula
x=−b±√Δ2ax=−b±√Δ2aDiscriminant Formula
Δ=b2−4acΔ=b2−4acFirst, compute for the discriminanty=3x2-4x-4y=3x2−4x−4a=3a=3 b=-4b=−4 c=-4c=−4ΔΔ == b2−4acb2−4ac Discriminant Formula == −42−4(3)(−4)−42−4(3)(−4) Substitute values == 16+4816+48 == 6464 Next, substitute the discriminant to the Quadratic Formula to find the xx interceptsxx == −b±√Δ2a−b±√Δ2a Quadratic Formula == −(−4)±√642(3)−(−4)±√642(3) Substitute values == 4±864±86 Write each root individuallyx1x1 == 4+864+86 == 126126 == 22 x1x1 == 4-864−86 == -46−46 == -23−23 Mark these 22 points on the xx axisFind the yy intercept by substituting x=0x=0yy == 3x2-4x-43x2−4x−4 == 3(0)2-4(0)-43(0)2−4(0)−4 Substitute x=0x=0 == 0-0-40−0−4 == -4−4 Mark this on the yy axisFinally, form the parabola by connecting the points -
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Question 5 of 7
5. Question
Graph using the discriminanty=2x2+3x-1y=2x2+3x−1-
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Quadratic Formula
x=−b±√Δ2ax=−b±√Δ2aDiscriminant Formula
Δ=b2−4acΔ=b2−4acFirst, compute for the discriminanty=2x2+3x-1y=2x2+3x−1a=2a=2 b=3b=3 c=-1c=−1ΔΔ == b2−4acb2−4ac Discriminant Formula == 32−4(2)(−1)32−4(2)(−1) Substitute values == 9+89+8 == 1717 Next, substitute the discriminant to the Quadratic Formula to find the xx interceptsxx == −b±√Δ2a−b±√Δ2a Quadratic Formula == −3±√172(2)−3±√172(2) Substitute values == -3±√174−3±√174 Write each root individuallyx1x1 == -3+√174−3+√174 == 1.12341.1234 == 0.2810.281 x1x1 == -3-√174−3−√174 == -7.1234−7.1234 == -1.781−1.781 Mark these 22 points on the xx axisFind the yy intercept by substituting x=0x=0yy == 2x2+3x-12x2+3x−1 == 2(0)2+3(0)-12(0)2+3(0)−1 Substitute x=0x=0 == 0+0-10+0−1 == -1−1 Mark this on the yy axisFinally, form the parabola by connecting the points -
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Question 6 of 7
6. Question
Graph using the discriminanty=x2-6x+9y=x2−6x+9-
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Quadratic Formula
x=−b±√Δ2ax=−b±√Δ2aAxis of Symmetry
x=−b2ax=−b2aDiscriminant Formula
Δ=b2−4acΔ=b2−4acFirst, find the axis of symmetryy=x2-6x+9y=x2−6x+9a=1a=1 b=-6b=−6 c=9c=9xx == −b2a−b2a Axis of Symmetry xx == −(−6)2(1)−(−6)2(1) Substitute values xx == 6262 xx == 33 Find where the graph touches the axis of symmetry by substituting x=3x=3 to the function. This would be the vertexyy == x2-6x+9x2−6x+9 == (3)2-6(3)+9(3)2−6(3)+9 Substitute x=3x=3 == 9-18+99−18+9 == 00 Hence, the vertex is at (3,0)(3,0)Next, compute for the discriminantΔΔ == b2−4acb2−4ac Discriminant Formula == −62−4(1)(9)−62−4(1)(9) Substitute values == 36-3636−36 == 00 Substitute the discriminant to the Quadratic Formula to find the xx interceptsxx == −b±√Δ2a−b±√Δ2a Quadratic Formula == −(−6)±√02(1)−(−6)±√02(1) Substitute values == 6±026±02 == 6262 == 33 There is only one root or xx intercept which is at (3,0)(3,0)Recall that the (3,0)(3,0) is also the vertexFind the yy intercept by substituting x=0x=0yy == x2-6x+9x2−6x+9 == (0)2-6(0)+9(0)2−6(0)+9 Substitute x=0x=0 == 0-0+90−0+9 == 99 Mark this on the yy axisFinally, form the parabola by connecting the points -
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Question 7 of 7
7. Question
Graph using the discriminanty=x2+2x-8y=x2+2x−8-
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Chapters- Chapters
Quadratic Formula
x=−b±√Δ2ax=−b±√Δ2aAxis of Symmetry
x=−b2ax=−b2aDiscriminant Formula
Δ=b2−4acΔ=b2−4acFirst, find the axis of symmetryy=x2+2x-8y=x2+2x−8a=1a=1 b=2b=2 c=-8c=−8xx == −b2a−b2a Axis of Symmetry xx == −22(1)−22(1) Substitute values xx == -22−22 xx == -1−1 Find where the graph touches the axis of symmetry by substituting x=3x=3 to the function. This would be the vertexyy == x2+2x-8x2+2x−8 == (-1)2+2(-1)-8(−1)2+2(−1)−8 Substitute x=-1x=−1 == 1-2-81−2−8 == -9−9 Hence, the vertex is at (-1,-9)(−1,−9)Next, compute for the discriminantΔΔ == b2−4acb2−4ac Discriminant Formula == −22−4(1)(−8)−22−4(1)(−8) Substitute values == 4+324+32 == 3636 Substitute the discriminant to the Quadratic Formula to find the xx interceptsxx == −b±√Δ2a−b±√Δ2a Quadratic Formula == −2±√362(1)−2±√362(1) Substitute values = -2±62 = -1±3 = -1-3,-1+3 x = -4,2 Finally, form the parabola by connecting the points -
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Quizzes
- Sum & Product of Roots 1
- Sum & Product of Roots 2
- Sum & Product of Roots 3
- Sum & Product of Roots 4
- Solving Equations by Factoring 1
- Solving Equations Using the Quadratic Formula
- Completing the Square 1
- Completing the Square 2
- Intro to Quadratic Functions (Parabolas) 1
- Intro to Quadratic Functions (Parabolas) 2
- Intro to Quadratic Functions (Parabolas) 3
- Graph Quadratic Functions in Standard Form 1
- Graph Quadratic Functions in Standard Form 2
- Graph Quadratic Functions by Completing the Square
- Graph Quadratic Functions in Vertex Form
- Write a Quadratic Equation from the Graph
- Write a Quadratic Equation Given the Vertex and Another Point
- Quadratic Inequalities 1
- Quadratic Inequalities 2
- Quadratics Word Problems 1
- Quadratics Word Problems 2
- Quadratic Identities
- Graphing Quadratics Using the Discriminant
- Positive and Negative Definite
- Applications of the Discriminant 1
- Applications of the Discriminant 2
- Solving Reducible Equations