Anonymous
Anonymous asked in Science & MathematicsMathematics · 3 weeks ago

Quadratic help, please?

Write the equation of a parabola that has a vertex of (-3, 1) and passes through the point (-2, -1)

7 Answers

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  • 3 weeks ago

    parabola | vertex (-3, 1)

    through (-2, -1) | Cartesian equation

    (y - y_0)^2 = 4 a (x - x_0)

    (assuming rotation angle 0°)

  • 3 weeks ago

    y = ax² + bx + c ← this is a parabola

    The parabola passes through the point (- 2 ; - 1)

    y = ax² + bx + c → when: x = - 2, then: y = - 1

    4a - 2b + c = - 1

    c = - 1 - 4a + 2b

    The parabola has a vertex of (- 3 ; 1), so the parabola passes through the point (- 3 ; 1)

    y = ax² + bx + c → when: x = - 3, then: y = 1

    9a - 3b + c = 1 → recall: c = - 1 - 4a + 2b

    9a - 3b - 1 - 4a + 2b = 1

    b = 5a - 2

    The parabola has a vertex of (- 3 ; 1), so you can say that: y'(- 3) = 0

    y' = 2ax + b ← this is the derivative → when: x = - 3, then: y' = 0

    - 6a + b = 0 → recall: b = 5a - 2

    - 6a + 5a - 2 = 0

    → a = - 2

    Recall: b = 5a - 2

    b = - 10 - 2

    → b = - 12

    Recall: c = - 1 - 4a + 2b

    c = - 1 + 8 - 24

    → c = - 17

    y = ax² + bx + c → y = - 2x² - 12x - 17

  • 3 weeks ago

    By inspection,

    (1) the parabola is open to the right, thus let

    a(y-1)^2=x+3

    =>

    a(-1-1)^2=-2+3

    =>

    a=1/4

    =>

    the equation is

    (y-1)^2=4(x+3).

    (2) the parabola is open downward, let

    y-1=a(x+3)^2

    =>

    -1-1=a(-2+3)^2

    =>

    -2=a

    =>

    the equation is

    y=1-2(x+3)^2.

  • ?
    Lv 7
    3 weeks ago

    If the parabola has vertex at (-3, 1) we can say:

    y = a(x + 3)² + 1

    As the parabola passes through point (-2, -1) we have:

    -1 = a(-2 + 3)² + 1

    so, -1 = a + 1

    Hence, a = -2

    so, y = -2(x + 3)² + 1

    Note: The line of symmetry is at x = -3...i.e. vertical

    We can have the case where the parabola has a horizontal line of symmetry.

    so, x = b(y - 1)² - 3

    Hence, with point (-2, -1) we get:

    -2 = b(-1 - 1)² - 3

    so, -2 = 4b - 3

    Hence, b = 1/4

    so, x = (1/4)(y - 1)² - 3

    or, (y - 1)² = 4x + 12

    :)>

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  • 3 weeks ago

    There are far more than two answers, but the simplest two are:

    y = (-1-1)/(-2+3)² (x+3)² + 1 which is y = -2(x+3)² + 1

    x = (-2+3)/(-1-1)² (y-1)² - 3 which is x = (1/4)(y-1)² - 3

    ————————————————————————

    The quadratic equation of the curve with the vertex at (x₀,y₀), with a line of reflection parallel to the y axis and passing through (x₁,y₁) is y = (y₁-y₀)/(x₁-x₀)² (x-x₀)² + y₀, while the quadratic equation of the curve with the same vertex and pass-through point, but with a line of reflection parallel to the x axis is x = (x₁-x₀)/(y₁-y₀)² (y-y₀)² + x₀.

  • 3 weeks ago

    Using the vertex form of parabola.

    given V(-3,1) and P(-2,-1)solving for ay = a(x - h)^2 + k

    - 1 = a[- 2 - (-3)]^2 + 1

    - 1 = a + 1

    - 1 - 1 = a

     - 2 = a

    a = -2

    The equation if the parabola is

    y = - 2(x + 3)^2 + 1 Answer//

  • y - 1 = a * (x - (-3))^2

    y - 1 = a * (x + 3)^2

    (y - 1) / (x + 3)^2 = a

    a = (-1 - 1) / (-2 + 3)^2

    a = (-2) / (1)^2

    a = -2/1

    a = -2

    y - 1 = -2 * (x + 3)^2

    y - 1 = -2 * (x^2 + 6x + 9)

    y - 1 = -2x^2 - 12x - 18

    y = -2x^2 - 12x - 17

    or

    x - (-3) = a * (y - 1)^2

    x + 3 = a * (y - 1)^2

    a = (x + 3) / (y - 1)^2

    a = (-2 + 3) / (-1 - 1)^2

    a = 1 / (-2)^2

    a = 1/4

    x + 3 = (1/4) * (y - 1)^2

    4x + 12 = (y - 1)^2

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