Given three points (-1,-7) (1, -5) (2, -1) how do you write a quadratic function in standard form with the points?

Question

Eva Adamson · Accepted Answer

Use the 3 points and the standard form

#y = ax^2+bx+c" [1]"#

to write 3 simultaneous equations with the variables, a, b , and c and then solve the equations.

Substitute the point #(-1,-7)# into equation [1]: #a(-1)^2+b(-1)+c = -7# #a - b+ c = -7" [2]"# Write equation [2] as the first row of an Augmented Matrix : #[ (1,-1,1,|,-7) ]# Substitute the point #(1,-5)# into equation [1]: #a(1)^2+b(1)+c = -5# #a + b+ c = -5" [3]"# Add a row corresponding to equation [3] to the augmented matrix: #[ (1,-1,1,|,-7), (1,1,1,|,-5) ]# Substitute the point #(2,-1)# into equation [1]: #a(2)^2+b(2)+c = -1# #4a + 2b+ c = -1" [4]"# Add a row corresponding to equation [4] to the augmented matrix: #[ (1,-1,1,|,-7), (1,1,1,|,-5), (4,2,1,|,-1) ]# Perform Elementary Row Operations on the matrix, until the left side becomes an Identity Matrix : #R_2-R_1toR_2# #[ (1,-1,1,|,-7), (0,2,0,|,2), (4,2,1,|,-1) ]# #R_3-4R_1toR_3# #[ (1,-1,1,|,-7), (0,2,0,|,2), (0,6,-3,|,27) ]# #1/2R_2toR_2# #[ (1,-1,1,|,-7), (0,1,0,|,1), (0,6,-3,|,27) ]# #R_1+R_2toR_1# #[ (1,0,1,|,-6), (0,1,0,|,1), (0,6,-3,|,27) ]# #R_3-6R_2toR_3# #[ (1,0,1,|,-6), (0,1,0,|,1), (0,0,-3,|,21) ]# #-1/3R_3 to R_3# #[ (1,0,1,|,-6), (0,1,0,|,1), (0,0,1,|,-7) ]# #R_1-R_3toR_1# #[ (1,0,0,|,1), (0,1,0,|,1), (0,0,1,|,-7) ]# We have an identity matrix on the left, therefore, the solution is the column vector on the right: #a = 1, b = 1 and c = -7# check #y = x^2+x-7" [5]"# by substituting in all 3 points. Substitute the point #(-1,-7)# into equation [5]: #-7=(-1)^2+ (-1)-7# #-7=-7# Substitute the point #(1,-5)# into equation [5]: #-5 = (1)^2+ 1 - 7# #-5=-5# Substitute the point #(2,-1)# into equation [5]: #-1= (2)^2+(2)-7# #-1= -1# This checks, therefore, equation [5] is the correct answer.

Jameson Allen · Answer

undefined To write a quadratic function in standard form using the given points $(-1,-7)$, $(1, -5)$, and $(2, -1)$, you can use the general form of a quadratic function $y = ax^2 + bx + c$ and substitute the coordinates of the points to form a system of equations. Then, solve the system to find the coefficients $a$, $b$, and $c$.

1. Substitute the coordinates of the first point $(-1,-7)$:
   $a(-1)^2 + b(-1) + c = -7$

2. Substitute the coordinates of the second point $(1, -5)$:
   $a(1)^2 + b(1) + c = -5$

3. Substitute the coordinates of the third point $(2, -1)$:
   $a(2)^2 + b(2) + c = -1$

Now, you have a system of three equations:

1. $a - b + c = -7$
2. $a + b + c = -5$
3. $4a + 2b + c = -1$

Solve this system of equations to find the values of $a$, $b$, and $c$, which will give you the quadratic function in standard form.