Find an equation in the form ax^2 +bx+c for the parabola passing through the points (-1,-15),(4,-105),(3,-55)

Answer 1

plug in each point into y=ax^2+bx+c to get 3 equations

(-1,-15)

-15 = a-b+c

(4,-105)

-105=16a+4b+c

(3,-55)

-55 = 9a +3b+c

3 equations 3 unknowns, solve by substitution elimination or matrix algebra

a=-8,

b=6

c=-1

y=f(x)=-8x^2 +6x -1 is the parabola

it helps to graph it and plot the 3 points to see if the answer looks reasonable

Answer 2

ax²+bx+c=y

From the 3 points

a-b+c=-15 eqn1

16a+4b+c=-105 eqn2

9a+3b+c=-55 eqn3

eqn2-4eqn1

20a+5c=-165 eqn4

eqn3-3eqn1

12a+4c=-100 --> 3a+c=-25 eqn5

eqn4-5eqn5

5a=-40 --> a = -8

substituting -8 for a in eqn5

c=-1

Substituting a and c into eqn1

-8-b-1=-15

b=6

y=-8x²+6x-1

Answer 3

Substituting the coordinates of the given points into the general equation for a parabola, results in:

(-1,-15)

I. a - b + c = -15

(4, -105)

II. 16a + 4b + c = -105

(3,-55)

III. 9a + 3b + c = -55

This is a system of three equations with three unknowns. Using addition elimination you can find the values for a, b, c as follows (there are several other methods for solving this system):

Try eliminating c from two different pairs of equations, perhaps by subtracting I. from II and I from III.

I from II:

15a + 5b = -90

IV. 3a + b = -18 (dividing every term by 5)

From I and III:

8a + 4b = -40

V. 2a + b = -10 (dividing every term by 4)

Now eliminate b from IV and V by subtracting V from IV.

a = -8

Use equation V to solve for b:

-16 + b = -10

b = 6

Use equation I to solve for c:

-8 - 6 + c = -15

c = -1

So the equation for the parabola is:

y = -8x² + 6x - 1

Note that the axis of symmetry is x = -b/2a = -6/2(-8) = 3/8

You can also find the x-intercepts by factoring (or quadratic formula)--left for you.

Check it out here:

desmos.com/calculator/jsj8c5ejcv