SOLUTION: Write the equation of a parabola, in standard form, that goes through these points: (0, 3) (1, 4) (-1, -6) Graph the parabola above. Indicate the vertex and axis

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Write the equation of a parabola, in standard form, that goes through these points: (0, 3) (1, 4) (-1, -6) Graph the parabola above. Indicate the vertex and axis       Log On


   

Question 887363: Write the equation of a parabola, in standard form, that goes through these points:
(0, 3) (1, 4) (-1, -6)

Graph the parabola above. Indicate the vertex and axis of symmetry.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
ax%5E2%2Bbx%2Bc=y

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0%2B0%2Bc=3
a%2A1%2Bb%2A1%2Bc=4
a%28-1%29%5E2%2Bb%28-1%29%2Bc=-6
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0%2B0%2Bc=3
a%2Bb%2Bc=4
a-b%2Bc=-6
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The variables in this now become a, b, c.
The y-intercept point and its equation allow eliminating c so you can form two equations in just the unknowns a and b.

You already see highlight%28c=3%29.

a%2Bb=4-c=4-3, highlight_green%28a%2Bb=1%29.

a-b=-6-c=-6-3, highlight_green%28a-b=-9%29.

Add those two and you can solve for a.
highlight%28a=-5%29.

Use either of the a & b equations to find b.
a%2Bb=1
b=1-a
b=1-%28-5%29
highlight%28b=6%29.

General equation now, highlight%28y=-5x%5E2%2B6x%2B3%29.
Is that factorable?
Discriminant, 6%2A6-4%2A%28-5%29%2A3=36%2B60=96, quadratic is NOT factorable.

You still want the vertex maximum point in order to be able to graph y.
You have your choice of completing the square to convert into standard form; or you can use the general solution of a quadratic equation to find the zeros and the vertex will be in the exact middle of the two solutions.

This may help: Completing the Square to Solve General Quadratic Equation

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

Write the equation of a parabola, in standard form, that goes through these points:
(0, 3) (1, 4) (-1, -6)

Graph the parabola above. Indicate the vertex and axis of symmetry.


Standard form of the quadratic equation: f%28x%29+=+-+4x%5E2+%2B+5x+%2B+3

Axis of symmetry is at: x+=+-+b%2F2a, or x+=+-+5%2F%282%28-+4%29%29, or x+=+%28-+5%29%2F-+8, or x+=+5%2F8

Substitute this value into the equation, and solve for k: the y-coordinate of the vertex. You will then 

have the coordinates of the vertex, or (h, k). This should help you to graph the parabola.