Write the equation, in standard form, of the parabola containing the following points:
(0, 1), (1, 5), (2, 3). You must set up a system of three equations to get credit for this question.
A. y= -3x^2+7x+1
B. y= -7x^2+3x+1
C. y= -x^2+5x+3
D. y= 3x^2-7x-1
i think B but idk 4 sure
first try (0,1) in each
Immediately it has to be A or B to get 1 when x = 0
now try (1,5 ) in A and B
-3+7 + 1 = 5 ok
-7 + 3 + 1 = -3 No way
Better be A try (2,3) inA
-3*4 + 14 +1
-12+ 14 +1 = 3 !!!! Caramba !!!!
Now the way they want
y = a x^2 + b x + c
(0,1) 1 = 0 + 0 + c so c = 1
so
y = a x^1 + b x + 1
(1,5) 5 = a + b + 1
so b = 4-a
so
y = a x^2 + (4-a) x + 1
(2,3) 3 =a *4 + (4-a) * 2 + 1
3 = 4 a + 8 - 2a + 1
-6 = 2a
a = -3
b = 4-a = 7
c = 1 What else is new :)
Well, solving this problem will be a bit like juggling, but don't worry, I'm an expert juggler-bot! Let's start by using the general form of a parabola equation: y = ax^2 + bx + c.
Using the first point (0, 1), we can substitute the values in the equation: 1 = a(0)^2 + b(0) + c. This simplifies to c = 1.
Using the second point (1, 5): 5 = a(1)^2 + b(1) + 1. This further simplifies to a + b = 4.
Lastly, using the third point (2, 3): 3 = a(2)^2 + b(2) + 1. This simplifies to 4a + 2b = 2.
So now we have our system of equations:
ⓐc = 1
ⓑa + b = 4
ⓒ4a + 2b = 2
Now let's solve this juggling act! Multiplying the equation ⓒ by 1/2, we get 2a + b = 1, which is easier to manipulate.
Subtracting the equation ⓑ from the equation 2a + b = 1, we have: 2a + b - (a + b) = 1 - 4. This simplifies to a = -3.
Substituting the value of a into the equation a + b = 4, we get -3 + b = 4. Solving for b gives us b = 7.
Plugging the values of a and b back into our parabola equation y = ax^2 + bx + c, we get y = -3x^2 + 7x + 1.
So, the correct answer is A. y = -3x^2 + 7x + 1.
To find the equation of a parabola in standard form, we can use the general form of the equation of a parabola: y = ax^2 + bx + c. Since we have three points, we can set up a system of three equations using the given points and then solve for the coefficients a, b, and c.
Let's use the first point (0, 1) to set up the first equation:
1 = a(0)^2 + b(0) + c
This simplifies to c = 1.
Now, let's use the second point (1, 5) to set up the second equation:
5 = a(1)^2 + b(1) + c
This simplifies to a + b + c = 5.
Finally, let's use the third point (2, 3) to set up the third equation:
3 = a(2)^2 + b(2) + c
This simplifies to 4a + 2b + c = 3.
Now we have the system of equations:
c = 1
a + b + c = 5
4a + 2b + c = 3
We can solve this system of equations to find the values of a, b, and c.
Substituting c = 1 into the second equation gives us:
a + b + 1 = 5
a + b = 4
Substituting c = 1 into the third equation gives us:
4a + 2b + 1 = 3
4a + 2b = 2
2a + b = 1
Now we have a system of two equations:
a + b = 4
2a + b = 1
We can solve this system using any method such as substitution or elimination. Let's use substitution:
From the first equation, we can express b in terms of a: b = 4 - a.
Substituting this expression for b into the second equation, we have:
2a + (4 - a) = 1
2a + 4 - a = 1
a + 4 = 1
a = -3
Substituting the value of a = -3 into the first equation, we have:
-3 + b = 4
b = 7
Now we have a = -3 and b = 7. Using c = 1, we have all the coefficients to write the equation in standard form:
y = -3x^2 + 7x + 1
Therefore, the correct equation in standard form is option A. y = -3x^2 + 7x + 1.