Find an equation of the parabola in standard form passing through the points: (-2, 9), (-4, 5), (1, 0)

Question

I tried to work the problem out myself and have gotten this:

y=ax^2+bx+c

(-2, 9), 4a-2b+c = 9

(-4, 5), 16a-4b+c = 5

(1, 0), a+b+c = 0

Then,

4a-2b+c = 9

a +b+c = 0

16a-4b+c = 5

a+b+c = 0

3a – b = 9

15a- 4b = 5

Answer 1

Hi Bo G

y = -x² - 4x + 5

You look like you are on the right path with three equations in three unknowns, its a long one but you can graph the equation above at Desmos.com to see that it passes through all the points given

Your parabola passes through (-4,5),(-2,9) and (1,0)

In Standard Form

y = ax² + bx + c

Those unknowns being a, b, c

for (1,0)

0 = a(1)² + b(1) + c

for (-2,9)

9 = a(-2)² + b(-2) + c

for (-4,5)

5 = a(-4)² + b(-4) + c

You three Equations are

Equation 1

a + b + c = 0

Equation 2

4a - 2b + c = 9

Equation 3

16a -4b + c = 5

We can reduce them to 2 Equations in 2 unknowns by Eliminating c

Multiply Equation 1 by negative 1

-1(a + b + c = 0)

-a - b -c = 0

Combine the above with Equation 2

-a - b - c = 0

4a - 2b +c = 9

This leaves

3a - 3b = 9

Next combine the adjusted Equation 1 with Equation 3 to eliminate c again

-a - b - c = 0

16a -4b + c = 5

This leaves

15a - 5b = 5

Now we 2 equations in 2 unknowns to work with

3a - 3b = 9

15a - 5b = 5

We can eliminate a if we multiply 3a - 3b =9 by negative 5

-5(3a - 3b = 9)

-15a + 15b = -45

Next we combine our adjusted equation with 15a - 5b = 5

-15a + 15b = -45

15a - 5b = 5

This leaves

10b = - 40

Divide both sides by 10

b = -4

Now go way back up to your original Equations insert b to get two Equations in a and c. Below I will use Equations 1 and 2

Equation 1

a + b + c = 0

a - 4 + c = 0

Add 4 to both sides of the equations

a + c = 4

Equation 2

4a - 2b + c = 9

4a - 2(-4) + c = 9

4a + 8 + c = 9

Subtract 8 from both sides of the equation

4a + c = 1

Now we can combine the two Equations in a and c to eliminate c

a + c = 4

4a + c = 1

Multiply a + c = 4 by negative 1

-1(a + c = 4)

-a - c = -4

Combine it with 4a + c = 1

-a - c = -4

4a + c = 1

This leaves

3a = -3

Divide both sides by 3

a = -1

Now use Equation 1 to solve for c

a + b + c =0

-1 - 4 + c = 0

-5 + c =0

Add 5 to both sides of the equation

c = 5

Plugging all the variables into

ax² + bx + c

y = -x² - 4x + 5

You can graph your parabola to make sure it passes through (-4,5)(-2,9)(1,0)

I hope this helps

Answer 2

it has be either a downward opening parabola of standard form y=a(x-h)^2 + k

or a rightward opening parabola of form x= a(y-h)^2 + k

plug in each points to get 3 equations, 2 unknowns. It may not even have a solution.

If it doesn't for the y=a(x-h)^2 + k form, then try the rightward opening parabola.

if neither works, you're into a less standard parabola that open at an angle with xy terms