what is the equation in standard form of a parabola that contain the following points?

(-2,-20),(0,-4),(4,-20)
Please Help

you can probably save some time by looking at the points. You know that parabolas are symmetric, so, since y(-2) = y(4), the axis of symmetry is at x=(-2+4)/2 = 1.

So, with x=1 the axis of symmetry (and hence at the vertex), you will have

y = a(x-1)^2+k
at x=0, y= -4, so
-4 = a+k
-20 = 9a + k

k=-2
a=-2

so, y = -2(x-1)^2 - 2
or, y = -2x^2 + 4x - 4

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or, if you let y = ax^2 + bx + c,

-20 = 4a - 2b + c
-4 = c
-20 = 16a + 4b + c

a = -2
b = 4
c = -4

y = -2x^2 + 4x - 4

Thanks for helping I have posted two other questions if you might be able to help

To find the equation in standard form of a parabola that passes through the given points, we can use the standard form equation of a parabola: y = ax^2 + bx + c.

Step 1: Plug in the coordinates (-2, -20)

-20 = a(-2)^2 + b(-2) + c
-20 = 4a - 2b + c

Step 2: Plug in the coordinates (0, -4)

-4 = a(0)^2 + b(0) + c
-4 = c

Step 3: Plug in the coordinates (4, -20)

-20 = a(4)^2 + b(4) + c
-20 = 16a + 4b + c

Since we know c = -4 (from step 2), we substitute this value into equations from step 1 and 3:

-20 = 4a - 2b - 4 (Equation 1)
-20 = 16a + 4b - 4 (Equation 2)

Simplifying Equation 1:

4a - 2b = -16 (Equation 1')

Simplifying Equation 2:

16a + 4b = -16 (Equation 2')

Step 4: Solve the system of equations (Equation 1' and Equation 2').

Multiply Equation 1' by 2:

8a - 4b = -32

Add this equation to Equation 2':

(8a - 4b) + (16a + 4b) = -32 + (-16)

Simplifying:

24a = -48

Divide by 24:

a = -48/24

a = -2

Substitute the value of a into Equation 1':

4(-2) - 2b = -16

-8 - 2b = -16

Subtract -8 from both sides:

-2b = -8

Divide by -2:

b = 4

Step 5: Write the equation in standard form.

Using the values of a = -2, b = 4, and c = -4 (from Step 2), the equation of the parabola in standard form is:

y = -2x^2 + 4x - 4

To find the equation in standard form of a parabola that passes through these three points, we can use the general form of a parabola equation: y = ax^2 + bx + c. We need to determine the values of a, b, and c.

Step 1: Plug in the given points into the equation to form a system of three equations:

For the point (-2, -20):
-20 = a(-2)^2 + b(-2) + c

For the point (0, -4):
-4 = a(0)^2 + b(0) + c

For the point (4, -20):
-20 = a(4)^2 + b(4) + c

Step 2: Simplify the equations:

For the point (-2, -20):
-20 = 4a - 2b + c

For the point (0, -4):
-4 = c

For the point (4, -20):
-20 = 16a + 4b + c

Step 3: Since we have obtained the value of c from the equation for the point (0, -4), we can substitute it into the other two equations:

For the point (-2, -20):
-20 = 4a - 2b - 4

For the point (4, -20):
-20 = 16a + 4b - 4

Step 4: Simplify further to eliminate c:

For the point (-2, -20):
-16 = 4a - 2b

For the point (4, -20):
-16 = 16a + 4b

Step 5: Solve the system of equations. There are multiple methods to solve this system, such as substitution or elimination. Let's use the elimination method:

Multiply the equation for the point (-2, -20) by 2 and add it to the equation for the point (4, -20) to eliminate b:

-32 = 8a - 4b
-16 = 16a + 4b
--------------
-48 = 24a

Solving for a:
a = -2

Substitute the value of a back into one of the previous equations to solve for b:

-16 = 16(-2) + 4b
-16 = -32 + 4b
16 = 4b
b = 4

Step 6: Substitute the values of a, b, and c back into the equation:

y = ax^2 + bx + c
y = -2x^2 + 4x - 4

Thus, the equation in standard form of the parabola passing through the given points (-2, -20), (0, -4), and (4, -20) is:

y = -2x^2 + 4x - 4