y= ax^2+bx-4. How would i find the equation of this parabola with the points (0,-4) (1,0) (2,2) (4,0) and 6,-10)?
Use the first few given points
for (0,-4) : -4 = 0 + 0 - 4 , nothing new learned here
for (1,0) : 0 = a + b - 4 ----> a+b+4 or b = 4-a
for (2,2) : 2 = 4a + 2b - 4 --- 2a + b = 3
use substitution
2a + (4-a) = 3
a = -1
then b = 4-(=1) = 5
y = -x^2 + 5x - 4
test if the other points satisfy, they do!
You must write equation for any pair of points.
y=ax^2+bx-4
x=0 y= -4
-4=a*0^2+b*0+c
0+0+c= -4
c= -4
x=1 y=0
0=a*1^2+b*1-4
a*1+b*1-4=0
a+b-4=0
a=4-b
x=2 y=2
y=ax^2+bx-4
2=(4-b)*2^2+b*2-4
(4-b)*4+2b-4=2
16-4b+2b=2+4
-2b=6-16
-2b= -1 Divide with -2
b=5
a=4-b
a=4-5
a= -1
So:
a= -1 b=5 c= -4
y=ax^2+bx+c= -x^2+5x-4
y=-x^2+5x-4
Checking of results:
For x=4
y=ax^2+bx-4
y= -4^2+5*4-4
y= -16+20-4
y=4-4
y=0
Correct value
For x=6
y=ax^2+bx-4
y= -6^2+6*5-4
y= -36+30-4
y= -6-4
y= -10
Correct value
I make one mistake in typig.
-2b= -1 Divide with -2 is mistake
-2b= -10 Divide with -2 is correct
To find the equation of the parabola with the given points, we need to substitute each point into the general form of a parabola equation, y = ax^2 + bx - 4, and solve the resulting system of equations.
Let's start by substituting the coordinates (0,-4) into the equation:
-4 = a(0)^2 + b(0) - 4
Simplifying, we have -4 = -4. This equation tells us nothing new about a or b.
Now, let's substitute the coordinates (1,0) into the equation:
0 = a(1)^2 + b(1) - 4
Simplifying, we have 0 = a + b - 4.
Next, let's substitute the coordinates (2,2) into the equation:
2 = a(2)^2 + b(2) - 4
Simplifying, we have 2 = 4a + 2b - 4.
Similarly, we substitute the coordinates (4,0):
0 = a(4)^2 + b(4) - 4
Simplifying, we have 0 = 16a + 4b - 4.
Lastly, we substitute the coordinates (6,-10):
-10 = a(6)^2 + b(6) - 4
Simplifying, we have -10 = 36a + 6b - 4.
Now we have a system of equations:
-4 = -4 --> Equation 1
0 = a + b - 4 --> Equation 2
2 = 4a + 2b - 4 --> Equation 3
0 = 16a + 4b - 4 --> Equation 4
-10 = 36a + 6b - 4 --> Equation 5
To solve this system of equations, we can either use substitution or elimination method. We'll use the elimination method for this example.
First, let's subtract Equation 3 from Equation 2:
0 - 2 = a + b - 4 - (4a + 2b - 4)
-2 = -3a - b + 4
Simplifying, we have -2 = -3a - b + 4.
Next, let's subtract Equation 4 from Equation 3:
0 - 0 = 4a + 2b - 4 - (16a + 4b - 4)
0 = -12a - 2b + 4
Simplifying, we have 0 = -12a - 2b + 4.
Lastly, let's subtract Equation 5 from Equation 3:
0 - (-10) = 4a + 2b - 4 - (36a + 6b - 4)
10 = -32a - 4b + 4
Simplifying, we have 10 = -32a - 4b + 4.
Now we have a new system of equations:
-2 = -3a - b + 4 --> Equation 6
0 = -12a - 2b + 4 --> Equation 7
10 = -32a - 4b + 4 --> Equation 8
To eliminate the b term, we can multiply Equation 6 by 2 and Equation 7 by 1, and then add them together:
-4 = -6a - 2b + 8 --> Equation 9
0 = -12a - 2b + 4 --> Equation 7
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-4 = -18a + 12 --> Equation 10
Next, we subtract Equation 7 from Equation 8:
10 = -32a - 4b + 4 --> Equation 8
0 = -12a - 2b + 4 --> Equation 7
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10 = -20a + 0 --> Equation 11
Now, we can solve Equation 10 for a:
-4 = -18a + 12
-18a = -16
a = -16/-18
a = 8/9
Using this value of a, we can substitute it into Equation 2 to solve for b:
0 = (8/9) + b - 4
b = 4 - (8/9)
b = 36/9 - 8/9
b = 28/9
Therefore, the equation of the parabola can be written as:
y = (8/9)x^2 + (28/9)x - 4