How would I go about solving this equation algebraically?
Find the equation of the parabola
y=ax^2 + bx+c
that passes through the given points.
(2,0) (3,-1) (4,0)
Thanks.
Sketch a graph.
see that vertex is at (3.-1)
so form is
(y+1) = k(x-3)^2
substitute in (2,0) to find k
1 = k (-1)^2
k = 1
so
y = (x-3)^2
or
y = x^2 - 6 x + 9
I plugged that into my graphing calculator and that isn't right. Could someone help?
To find the equation of the parabola that passes through the given points, we can plug in the x and y coordinates of each point into the equation of a parabola, and then solve the resulting system of equations for the variables a, b, and c.
Let's start by plugging in the coordinates of the first point (2,0) into the equation:
0 = a(2^2) + b(2) + c
Simplifying this equation gives us:
4a + 2b + c = 0 --(1)
Now, let's do the same for the second point (3,-1):
-1 = a(3^2) + b(3) + c
Simplifying this equation gives us:
9a + 3b + c = -1 --(2)
Lastly, let's plug in the coordinates of the third point (4,0) into the equation:
0 = a(4^2) + b(4) + c
Simplifying this equation gives us:
16a + 4b + c = 0 --(3)
Now we have a system of three equations (equations 1, 2, and 3) with three variables (a, b, and c). We can solve this system of equations to find the values of a, b, and c that satisfy all three equations.
There are multiple methods to solve this system, such as substitution or elimination. I will use the method of substitution to solve this system.
First, let's solve equation 1 for c:
c = -4a - 2b --(4)
Next, let's solve equation 2 for c:
c = -9a - 3b - 1 --(5)
Now, we can substitute equation 4 into equation 3 to eliminate c:
16a + 4b + (-4a - 2b) = 0
Simplifying this equation gives us:
12a + 2b = 0 --(6)
Similarly, we can substitute equation 5 into equation 3 to eliminate c:
16a + 4b + (-9a - 3b - 1) = 0
Simplifying this equation gives us:
7a + b = 1 --(7)
Now we have a system of two equations (equations 6 and 7) with two variables (a and b). We can solve this system to find the values of a and b.
To solve this system, let's multiply equation 6 by 7:
84a + 14b = 0 --(8)
Subtracting equation 7 from equation 8 will eliminate b:
(84a + 14b) - (7a + b) = 0 - 1
Simplifying this equation gives us:
77a + 13b = -1 --(9)
Now, let's solve equation 9 for b:
13b = -1 - 77a
b = (-1 - 77a) / 13
b = (-1/13) - (77/13)a --(10)
Now that we have the value of b, we can substitute it back into equation 6 to solve for a:
12a + 2((-1/13) - (77/13)a) = 0
Simplifying this equation gives us:
12a + ((-2)/13) - (154/13)a = 0
Subtracting ((-2)/13) from both sides and combining like terms gives us:
12a - (154/13)a = (2/13)
Simplifying this equation gives us:
(138/13)a = (2/13)
Dividing both sides by (138/13) gives us:
a = (2/13) / (138/13)
Simplifying this equation gives us:
a = 2/138
Simplifying further, we get:
a = 1/69
Now that we have the value of a, we can substitute it back into equation 10 to solve for b:
b = (-1/13) - (77/13)(1/69)
Simplifying this equation gives us:
b = (-1/13) - (77/13) * (1/69)
b = (-1/13) - (77/897)
b = (-1/13) - (77/897)
b = (-897 - 10081) / 897
b = -10978 / 897
Therefore, the equation of the parabola that passes through the given points (2,0), (3,-1), and (4,0) is:
y = (1/69)x^2 - (10978/897)x + c
To find the value of c, we can substitute the x and y coordinates of any of the given points into the equation and solve for c. Let's use the point (2,0):
0 = (1/69)(2^2) - (10978/897)(2) + c
Simplifying this equation gives us:
0 = (4/69) - (21956/897) + c
Combining like terms gives us:
0 = (4668 - 21956 + 69c) / 897
Simplifying this equation further gives us:
0 = (-17288 + 69c) / 897
Multiplying both sides by 897 gives us:
0 = -17288 + 69c
Adding 17288 to both sides gives us:
17288 = 69c
Dividing both sides by 69 gives us:
c = 17288 / 69
Therefore, the equation of the parabola that passes through the given points (2,0), (3,-1), and (4,0) is:
y = (1/69)x^2 - (10978/897)x + 17288/69