# 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

- KrishnamurthyLv 73 weeks ago
parabola | vertex (-3, 1)

through (-2, -1) | Cartesian equation

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

(assuming rotation angle 0°)

- la consoleLv 73 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

- PinkgreenLv 73 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 73 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

:)>

- What do you think of the answers? You can sign in to give your opinion on the answer.
- Φ² = Φ+1Lv 73 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.

- Engr. RonaldLv 73 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//

- 3 weeks ago
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