How do you find a quadratic function that passes through the points #(1, 4)#, #(-1, 6)# and #(-2, 16)# ?

Question

Avery Alexander · Accepted Answer

Function is #y=3x^2-x+2# Let the quadratic function be #y=ax^2+bx+c#. As it passes through #(1,4)#, #(-1,6)# and #-2,16#, we will have #a(1)^2+b*1+c=4# i.e. #a+b+c=4# ................................(1) #a(-1)^2+b*(-1)+c=6# i.e. #a-b+c=6# ................................(2) #a(-2)^2+b*(-2)+c=6# i.e. #4a-2b+c=16# ................................(3) Adding (1) and (2) gives us #2a+2c=10# i.e. #a+c=5# and #b=-1#................................(4) Multiplying (1) by #2# and adding it to (3), we get #2a+2b+2c+4a-2b+c=2xx4+16# or #6a+3c=24# i.e. #2a+c=8# ................................(5) Subtracting (4) from (5) #a=3# and hence #c=2# and function is #y=3x^2-x+2# graph{3x^2-x+2 [-3, 3, -2, 18]}

Abigail Allen · Answer

#f(x) = 3x^2-x+2#

Given:

#(1, 4)#, #(-1, 6), (-2, 16)#

We can immediately write down a formula for a quadratic that goes through these points by constructing terms for each distinct value of #x# we want to match:

#f(x) = color(blue)(4)*((x-(color(blue)(-1)))(x-(color(blue)(-2))))/(((color(blue)(1))-(color(blue)(-1)))((color(blue)(1))-(color(blue)(-2)))) +#

#color(white)(f(x) =) color(blue)(6)*((x-(color(blue)(1)))(x-(color(blue)(-2))))/(((color(blue)(-1))-(color(blue)(1)))((color(blue)(-1))-(color(blue)(-2)))) +#

#color(white)(f(x) =) color(blue)(16)*((x-(color(blue)(1)))(x-(color(blue)(-1))))/(((color(blue)(-2))-(color(blue)(1)))((color(blue)(-2))-(color(blue)(-1))))#

#color(white)(f(x)) = color(blue)(4) * ((x+1)(x+2))/((2)(3)) + color(blue)(6) * ((x-1)(x+2))/((-2)(1)) + color(blue)(16) * ((x-1)(x+1))/((-3)(-1))#

#color(white)(f(x)) = 2/3 (x^2+3x+2) - 3 (x^2+x-2) + 16/3(x^2-1)#

#color(white)(f(x)) = 3x^2-x+2#

This works by adding together the desired multiples of quadratics that take the value #1# at a given #x# coordinate and #0# at the others.

graph{(y-(3x^2-x+2))(8(x-1)^2+(y-4)^2-0.02)(8(x+1)^2+(y-6)^2-0.02)(8(x+2)^2+(y-16)^2-0.02)=0 [-5.35, 4.7, -3, 20]}

In general, given three points #(x_1, y_1)#, #(x_2, y_2)# and #(x_3, y_3)# with #x_1, x_2, x_3# all distinct, a quadratic function through them is given by:

#f(x) = y_1 ((x-x_2)(x-x_3))/((x_1-x_2)(x_1-x_3)) + y_2 ((x-x_1)(x-x_3))/((x_2-x_1)(x_2-x_3)) + y_3 ((x-x_1)(x-x_2))/((x_3-x_1)(x_3-x_2))#

Ethan Adkins · Answer

undefined To find a quadratic function that passes through the points (1, 4), (-1, 6), and (-2, 16):

1. Substitute the x and y coordinates of each point into the quadratic function equation $y = ax^2 + bx + c$.
2. You will get a system of three equations with three unknowns (a, b, and c).
3. Solve the system of equations to find the values of a, b, and c.
4. Once you have the values of a, b, and c, plug them back into the quadratic function equation to get the specific function.

Would you like further clarification on any of these steps?