How to put a Quadratic Graph into Standard Form (ax^2+bx+c)?

Question

I have a two parabolas... Help me create a equation in Standard Form for both. 

1) - Vertex: (-2,10) 
- Y Intercept: (0, 9) 
- X Intercept: (-10, 0) and (6,0) 

2) - Vertex: (-3.5, 6) 
- Y Intercept: (0, 5) 
- X Intercept: (-10, 0) and (3, 0)

Answer 1

The standard form of the quadratic function is f(x) = ax^2 + bx + c .

• 1).

If the x - intercepts are h and k, then the parabola equation is y = a(x - h)(x - k), where a is a is a constant.

The x - intercepts are (- 10, 0) and (6, 0).

So, the parabola equation is y = a(x - (- 10))(x - 6)

y = a(x + 10)(x - 6)

The y - intercept is (0, 9).

Substitute the values of (x, y) = (0, 9) in y = a(x + 10)(x - 6).

9 = a(0 + 10)(0 - 6)

9 = a * 10 * (- 6)

a = - 9/60

⇒ a = - 3/20.

Substitute a = - 3/20 in y = a(x + 10)(x - 6).

y = (- 3/20)[ (x + 10)(x - 6) ]

y = (- 3/20(x^2 + 4x - 60)

⇒ y = (- 3/20)x^2 - (3/5)x + 9.

The standard form of the parabola equation is y = (- 3/20)x^2 - (3/5)x + 9.

• 2).

If the x - intercepts are h and k, then the parabola equation is y = a(x - h)(x - k), where a is a is a constant.

The x - intercepts are (- 10, 0) and (3, 0).

So, the parabola equation is y = a(x - (- 10))(x - 3)

y = a(x + 10)(x - 3)

The y - intercept is (0, 5).

Substitute the values of (x, y) = (0, 5) in y = a(x + 10)(x - 3).

5 = a(0 + 10)(0 - 3)

5 = a * 10 * (- 3)

a = - 5/30

⇒ a = - 1/6.

Substitute a = - 1/6 in y = a(x + 10)(x - 3).

y = (- 1/6)[ (x + 10)(x - 3) ]

y = (- 1/6(x^2 + 7x - 30)

⇒ y = (- 1/6)x^2 - (7/6)x + 5.

The standard form of the parabola equation is y = (- 1/6)x^2 - (7/6)x + 5.

Comments

1)The parabola y = (- 3/20)x^2 - (3/5)x + 9 make the vertex (- 2, 9.6) so there is no parabola that fit all 4 of those points.

 

2)The parabola y = (- 1/6)x^2 - (7/6)x + 5 make the vertex (- 3.5, 7.04167) so there is no parabola that fit all 4 of those points.
 

Answer 2

1) y = a(x + 10)(x − 6)
=> y = a(x^2 + 4x − 60)
given the vertex (-2,10)
=> 10 = a((-2)^2 + 4 *-2 − 60)
=> 10 = -64a
=> a = -10/64
=> a = -5/32

=> y = -(5/32)x^2 −(5/8)x + (75/8)
however that make the y-intercept (0,9.375) so there is no parabola that fit all 4 of those points.

2) y = a(x + 10)(x − 3)
=> y = a(x^2 + 7x − 30)
given the vertex (-3.5,6)
=> 6 = a((-7/2)^2 + 7 * (-7/2) − 30)
=> 6 = -169a/4
=> -24 = 169a
=> a = -24/169

=> y = -(24/169)x^2 − (168/169)x + 720/169
this make the y intercept ≈4.26 so there is no parabola that fit all 4 of those points.

Source : https://answers.yahoo.com/question/index?qid=20140512183605AAfIEzF