How do you find a quadratic function that passes through the points #(1, 4)#, #(-1, 6)# and #(-2, 16)# ?
2 Answers
Function is
Explanation:
Let the quadratic function be
Adding (1) and (2) gives us
i.e.
Multiplying (1) by
or
Subtracting (4) from (5)
and function is
graph{3x^2-x+2 [-3, 3, -2, 18]}
Explanation:
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
#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
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
#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))#