A quadratic curve passes through the points (-2,0)and(1,0).Find the equation of the curve in the form ax^2+bx+c,where a,b and c are constants
(x+2)(x-1) = 0
thanku
To find the equation of the quadratic curve, we can use the fact that it passes through the points (-2,0) and (1,0).
Let's start by considering the general form of a quadratic equation: y = ax^2 + bx + c.
Since the curve passes through (-2,0), we can substitute these values into our equation to get:
0 = a(-2)^2 + b(-2) + c
Simplifying this equation, we get:
0 = 4a - 2b + c
Similarly, since the curve passes through (1,0), we substitute these values into our equation to get:
0 = a(1)^2 + b(1) + c
Simplifying this equation, we get:
0 = a + b + c
Now we have a system of equations with three variables (a, b, and c). We can solve this system to find their values.
Subtracting the second equation from the first equation, we get:
0 = 3a - 3b
This can be further simplified to:
a = b
Now we substitute this value of a into one of the equations to find the value of c. Let's substitute it into the second equation:
0 = b + b + c
0 = 2b + c
Since a = b, we can substitute a for b in the equation above:
0 = 2a + c
Now we have two equations: a = b and 0 = 2a + c.
We can solve this system of equations to find the values of a, b and c. Substituting a = b into 0 = 2a + c, we get:
0 = 2(a) + c
0 = 2a + c
Since we know a = b, we can write:
0 = 2b + c
We now have two equations:
a = b
0 = 2b + c
We can choose any value for a (or b) and solve the system to find the corresponding values for b (or a) and c. Let's choose a = 1:
From the equation a = b, we get b = 1.
Substituting this into the equation 0 = 2b + c, we get:
0 = 2(1) + c
0 = 2 + c
Subtracting 2 from both sides, we find:
c = -2
So, we have a = 1, b = 1, and c = -2.
Therefore, the equation of the quadratic curve in the form ax^2 + bx + c is:
y = x^2 + x - 2
To find the equation of a quadratic curve in the form ax^2 + bx + c, using the given points (-2,0) and (1,0), we can follow these steps:
Step 1: Plug in the x and y coordinates of each point into the equation to form two equations:
For the point (-2,0):
0 = a(-2)^2 + b(-2) + c --> 0 = 4a - 2b + c
For the point (1,0):
0 = a(1)^2 + b(1) + c --> 0 = a + b + c
These equations will help us solve for the values of a, b, and c.
Step 2: Solve the system of equations to find the values of a, b, and c:
From the equations:
0 = 4a - 2b + c
0 = a + b + c
We can use substitution or elimination method to solve for the variables. Let's use the elimination method:
Multiply the second equation by -4 to eliminate the 'a' terms:
0 = 4a - 2b + c
0 = -4a - 4b - 4c
Add the two equations together:
0 = 4a - 2b + c
0 = -4a - 4b - 4c
------------------------
0 = -6b - 3c
Simplify the equation:
0 = -6b - 3c --> 0 = 6b + 3c
Now, we have one equation with two variables (b and c). We can plug in any value for 'b' or 'c' to solve for the other variable.
Let's assume 'b' to be 1:
0 = 6(1) + 3c --> 0 = 6 + 3c
Rearrange to solve for 'c':
-6 = 3c
Divide by 3:
-2 = c
Now, we have found the value of 'c' as -2. So, 'c' = -2.
Step 3: Substitute the value of 'c' into one of the original equations to solve for 'a' or 'b'.
Let's use the equation: 0 = a + b + c
0 = a + b - 2
Rearrange the equation:
2 = a + b
Now, we have one equation with two variables (a and b). We can plug in any value for 'a' or 'b' to solve for the other variable.
Let's assume 'b' to be 1:
2 = a + (1)
Subtract 1 from both sides:
1 = a
Now, we have found the value of 'a' as 1. So, 'a' = 1.
Step 4: Substitute the values of 'a', 'b', and 'c' into the equation ax^2 + bx + c to get the final equation of the quadratic curve:
The equation of the curve is: 1x^2 + 1x - 2
Simplifying it further:
Final equation: x^2 + x - 2